The general formula for the amount in savings account compounded annually is given as;
[tex]\begin{gathered} A=P(1+\frac{r}{100n})^{nt} \\ \text{Where A=Amount} \\ P=\text{Initial deposit} \\ r=\text{rate} \\ n=n\text{ umber of times it is compounded annually} \\ t=\text{time} \end{gathered}[/tex]A. The equation for the value of the investment as a function of t in the current account they have is;
[tex]A(t)=\text{ \$30000(1+}\frac{5.25}{100})^t[/tex]B. The equation for the value of the investment in an account earning 5.875% interest compounded annually is;
[tex]A(t)=\text{ \$30000(1+}\frac{5.875}{100})^{t^{}}[/tex]C. The equation for the value of the investment in an account earning 5.75% compounded semi-annually; that is twice in a year is;
[tex]\begin{gathered} A(t)=\text{ \$30000(1+}\frac{5.75}{100(2)})^{2t} \\ A(t)=\text{ \$30000(1+}\frac{5.75}{200})^{2t} \end{gathered}[/tex]D. The solution for the value of the investment in an account earning 5.5% annual interest compounded quarterly; that is four times in a year;
[tex]\begin{gathered} A(t)=\text{ \$30000(1+}\frac{5.5}{100(4)})^{4t} \\ A(t)=\text{ \$30000(1+}\frac{5.5}{400})^{4t} \end{gathered}[/tex]let f(x)=8x+5 and g(x)=9x-2. find the function.f - g(f - g) (x) =find the domain.
Answer:
(f - g)( x ) = -x + 7
Domain;
[tex](-\infty,\infty)[/tex]Explanation:
Given the below functions;
[tex]\begin{gathered} f(x)=8x+5 \\ g(x)=9x-2 \end{gathered}[/tex]To find (f - g)( x ), all we need to do is subtract g(x) from f(x) as shown below;
[tex]\begin{gathered} (f-g)\mleft(x\mright)=(8x+5)-(9x-2) \\ =8x+5-9x+2 \\ =8x-9x+5+2 \\ =-x+7 \end{gathered}[/tex]The domain of the function will be all values from negative infinity to positive infinty, written as;
[tex](-\infty,\infty)[/tex]Number 14. Directions in pic. And also when you graph do the main function in red and the inverse in blue
Question 14.
Given the function:
[tex]f(x)=-\frac{2}{3}x-4[/tex]Let's find the inverse of the function.
To find the inverse, take the following steps.
Step 1.
Rewrite f(x) for y
[tex]y=-\frac{2}{3}x-4[/tex]Step 2.
Interchange the variables:
[tex]x=-\frac{2}{3}y-4[/tex]Step 3.
Solve for y
Add 4 to both sides:
[tex]\begin{gathered} x+4=-\frac{2}{3}y-4+4 \\ \\ x+4=-\frac{2}{3}y \end{gathered}[/tex]Multply all terms by 3:
[tex]\begin{gathered} 3x+3(4)=-\frac{2}{3}y\ast3 \\ \\ 3x+12=-2y \end{gathered}[/tex]Divide all terms by -2:
[tex]\begin{gathered} -\frac{3}{2}x+\frac{12}{-2}=\frac{-2y}{-2} \\ \\ -\frac{3}{2}x-6=y \\ \\ y=-\frac{3}{2}x-6 \end{gathered}[/tex]Therefore, the inverse of the function is:
[tex]f^{-1}(x)=-\frac{3}{2}x-6[/tex]Let's graph both functions.
To graph each function let's use two points for each.
• Main function:
Find two point usnig the function.
When x = 3:
[tex]\begin{gathered} f(3)=-\frac{2}{3}\ast3-4 \\ \\ f(3)=-2-4 \\ \\ f(3)=-6 \end{gathered}[/tex]When x = 0:
[tex]\begin{gathered} f(0)=-\frac{2}{3}\ast(0)-4 \\ \\ f(-3)=-4 \end{gathered}[/tex]For the main function, we have the points:
(3, -6) and (0, -4)
Inverse function:
When x = 2:
[tex]\begin{gathered} f^{-1}(2)=-\frac{3}{2}\ast(2)-6 \\ \\ f^{-1}(2)=-3-6 \\ \\ f^1(2)=-9 \end{gathered}[/tex]When x = -2:
[tex]\begin{gathered} f^{-1}(-2)=-\frac{3}{2}\ast(-2)-6 \\ \\ f^1(-2)=3-6 \\ \\ f^{-1}(2)=-3 \end{gathered}[/tex]For the inverse function, we have the points:
(2, -9) and (-2, -3)
To graph both functions, we have:
ANSWER:
[tex]\begin{gathered} \text{ Inverse function:} \\ f^{-1}(x)=-\frac{3}{2}x-6 \end{gathered}[/tex]Which comparison is NOT correct?2 > -3-7 < -5-9 < 10 < -4
0 > -4 is incorrect
as -4 is a negative number and it comes on the left of 0 on a number line
and we know number increase from left to right
so option D is the answer.
Consider the following loan. Complete parts (a)-(c) below.An individual borrowed $67,000 at an APR of 3%, which will be paid off with monthly payments of 347$ for 22 years.a. Identify the amount borrowed, the annual interest rate, the number of payments per year, the loan term, and the payment amount.The amount borrowed is $____ the annual interest rate is ____, the number of payments per year is _____, the loan term is _____ years, and the payment amount is _____$ b. How many total payments does the loan require? What is the total amount paid over the full term of the loan?There are ____ payments toward the loan and the total amount paid is ____$ c. Of the total amount paid, what percentage is paid toward the principal and what percentage is paid for interest?The percentage paid toward the principal is _____% and the percentage paid for interest is ____%.(Round to the nearest tenth as needed.)
b) There are 12 payments per year for 22 years; multiply 12 by 22 to get the total number of payments:
[tex]12\times22=264[/tex]To find the total amount paid, multiply the number of payments by the payment amount:
[tex]264\times347=91,608[/tex]There are 264 payments toward the loan and the total amount paid is $91,608c) Toward principal: $67,000
Toward interest: subtract the principal from the payment amount:
[tex]91,608-67,000=24,608[/tex]Let 91,608 be the 100%, use a rule of three to find the % corresponding to the principal and interest:
[tex]\begin{gathered} Principal: \\ x=\frac{67,000\times100}{91,608}=73.1 \\ \\ Interest: \\ x=\frac{24,608\times100}{91,608}=26.9 \end{gathered}[/tex]The percentage paid toward the principal is 73.1% and the percentage paid for interest is 26.9%The width of a rectangle is 6 less than twice its length. If the area of the rectangle is 170 cm2 , what is the length of the diagonal?The length of the diagonal is cm.Give your answer to 2 decimal places.Submit QuestionQuestion 25
The formula to find the area of a rectangle is:
[tex]\begin{gathered} A=l\cdot w \\ \text{ Where} \\ \text{ A is the area} \\ l\text{ is the length} \\ w\text{ is the width} \end{gathered}[/tex]Since the rectangle area is 170cm², we can write the following equation.
[tex]170=l\cdot w\Rightarrow\text{ Equation 1}[/tex]On the other hand, we know that the width of the rectangle is 6 less than twice its length. Then, we can write another equation.
[tex]\begin{gathered} w=2l-6\Rightarrow\text{ Equation 2} \\ \text{ Because} \\ 2l\Rightarrow\text{ Twice length} \\ 2l-6\Rightarrow\text{ 6 less than twice length} \end{gathered}[/tex]Now, we solve the found system of equations.
[tex]\begin{cases}170=l\cdot w\Rightarrow\text{ Equation 1} \\ w=2l-6\Rightarrow\text{ Equation 2}\end{cases}[/tex]For this, we can use the substitution method.
Step 1: we replace the value of w from Equation 2 into Equation 1. Then, we solve for l.
[tex]\begin{gathered} 170=l(2l-6) \\ \text{Apply the distributive property} \\ 170=l\cdot2l-l\cdot6 \\ 170=2l^2-6l \\ \text{ Subtract 170 from both sides} \\ 0=2l^2-6l-170 \end{gathered}[/tex]We can use the quadratic formula to solve the above equation.
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}\Rightarrow\text{ Quadratic formula} \\ \text{ For }ax^2+bx+c=0 \end{gathered}[/tex]Then, we have:
[tex]\begin{gathered} a=2 \\ b=-6 \\ c=-170 \\ l=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ l=\frac{-(-6)\pm\sqrt[]{(-6)^2-4(2)(-170)}}{2(2)} \\ l=\frac{6\pm\sqrt[]{1396}}{4} \\ \end{gathered}[/tex]There are two solutions for l.
[tex]\begin{gathered} l_1=\frac{6+\sqrt[]{1396}}{4}\approx10.84 \\ l_2=\frac{6-\sqrt[]{1396}}{4}\approx-7.84 \\ \text{ The symbol }\approx\text{ is read 'approximately'.} \end{gathered}[/tex]Since the value of l can not be negative, the value of l is 10.84.
Step 2: We replace the value of l into any of the equations of the system to find the value of w. For example, in Equation 1.
[tex]\begin{gathered} 170=l\cdot w\Rightarrow\text{ Equation 1} \\ 170=10.84\cdot w \\ \text{ Divide by 10.84 from both sides} \\ \frac{170}{10.84}=\frac{10.84\cdot w}{10.84} \\ 15.68\approx w \end{gathered}[/tex]Now, the long side, the wide side and the diagonal of the rectangle form a right triangle.
Then, we can use the Pythagorean theorem formula to find the length of the diagonal.
[tex]\begin{gathered} a^2+b^2=c^2 \\ \text{ Where} \\ a\text{ and }b\text{ are the legs} \\ c\text{ is the hypotenuse} \end{gathered}[/tex]In this case, we have:
[tex]\begin{gathered} a=10.84 \\ b=15.68 \\ a^2+b^2=c^2 \\ (10.84)^2+(15.68)^2=c^2 \\ 117.51+245.86=c^2 \\ 363.37=c^2 \\ \text{ Apply square root to both sides of the equation} \\ \sqrt[]{363.37}=\sqrt[]{c^2} \\ 19.06=c \end{gathered}[/tex]Therefore, the length of the diagonal of the given rectangle is 19.06 cm rounded to 2 decimal places.
if f(x)=-2x-3, find f(-1)
Solve;
[tex]\begin{gathered} f(x)=-2x-3 \\ f(-1)=-2(-1)-3 \\ f(-1)=2-3 \\ f(-1)=-1 \end{gathered}[/tex]The answer is -1
That is f(-1) = -1
The value of an IBM share one day was $ 74.50 more than the value of an AT&T share.
An algebraic expression we can use to compare the price of IBM shares as being $74.50 more than AT&T shares is x + 74.50, where x is the value of AT&T shares.
What is an algebraic expression?An algebraic expression consists of variables, terms, constants, and mathematical operations, including addition, subtraction, multiplication, division, and others.
The five algebraic expressions include monomial, polynomial, binomial, trinomial, multinomial.
We can also describe algebraic expressions as falling under the following categories:
Elementary algebraAdvanced algebraAbstract algebraLinear algebraCommutative algebra.An example of an algebraic expression is 2x + 3y.
Let the value of AT&T share = x
Let the value of IBM share = x + 74.50
Thus, we can, algebraically, conclude that AT&T's share price is x while the price of IBM's share is x + 74.50 on that particular day.
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Find the prime factorization of the following number write any repeated factors using exponents
Notice that 100=10*10, and 10=2*5. 2 and 5 are prime numbers; therefore,
[tex]\begin{gathered} 100=10\cdot10=(2\cdot5)(2\cdot5)=2\cdot2\cdot5\cdot5=2^2\cdot5^2 \\ \Rightarrow100=2^2\cdot5^2 \end{gathered}[/tex]The answer is 100=2^2*5^2
Write an addition equation and a subtraction equation
to represent the problem using? for the unknown.
Then solve.
There are 30 actors in a school play. There are
10 actors from second grade. The rest are from third
grade. How many actors are from third grade?
a. Equations:
b. Solve
The Equation is 10 + x= 30 and 20 actors are from third grade.
What is Equation?Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side. It demonstrates the equality of the relationship between the expressions printed on the left and right sides. LHS = RHS is a common mathematical formula.
Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.
Given:
There are 30 actors in a school play.
There are 10 actors from second grade.
The rest are from third grade.
let the actors in third grade is x.
Equation is:
Actors from second grade + Actors from third grade = Total actors
10 + x= 30
Now, solving
Subtract 10 from both side
10 +x - 10 = 30 - 10
x = 20
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What are all of the answers for these questions? Use 3 for pi. Please do not use a file to answer, I cannot read it.
The company's sign has two(2) congruent trapezoids and two(2) congruent right angled triangle.
The area of the figure is:
[tex]A_{\text{figure}}=2A_{\text{trapezoid}}+2A_{\text{triangle}}[/tex]The area of a trapezoid is given by the formula:
[tex]\begin{gathered} A_{\text{trapezoid}}=\frac{1}{2}(a+b)h \\ \text{where a and b are opposite sides of the trapezoid} \\ h\text{ is the height} \end{gathered}[/tex]Thus, we have:
[tex]\begin{gathered} A_{\text{trapezoid}}=\frac{1}{2}(1\frac{1}{2}+3)2 \\ A_{\text{trapezoid}}=\frac{1}{2}(1.5+3)2 \\ A_{\text{trapezoid}}=\frac{1}{2}\times4.5\times2=4.5m^2 \end{gathered}[/tex]Area of a triangle is given by the formula:
[tex]A_{\text{triangle}}=\frac{1}{2}\times base\times height[/tex]Thus, we have:
[tex]\begin{gathered} A_{\text{triangle}}=\frac{1}{2}\times2\times1\frac{1}{2} \\ A_{\text{triangle}}=\frac{1}{2}\times2\times1.5=1.5m^2 \end{gathered}[/tex]Hence, the area of the company's sign is:
[tex]\begin{gathered} A=(2\times4.5)+(2\times1.5) \\ A=9+3=12m^2 \end{gathered}[/tex]what's the answer for proportions 7/9=b/b-10
Answer:
-35
Step-by-step explanation:
[tex]\frac{7}{9}[/tex] = [tex]\frac{b}{b - 10}[/tex] multiply both sides by 9(b -10)
[tex]\frac{9(b - 10)}{1}[/tex] [tex](\frac{7}{9})[/tex] = [tex]\frac{9(b -10)}{1}[/tex] [tex](\frac{b}{b-10})[/tex] On the right side of the equation, the 9's cancel out and on the right side of the equation the (b -10) cancels out to leave
7(b -10) = 9b Distribute the 7
7b - 70 = 9b Subtract 7b from both sides
-70 = 2b Divide both sides by 2
-35 = b
how do you find the domain in a range of number 2?
The domain is all the x values included in the function, while the range are all the y values included in the function.
Based on the graph:
Answer:
• Domain:
[tex](-\infty,\text{ }\infty)[/tex]• Range:
[tex](0,\infty)[/tex]Kindly help by providing answers to these questions.
Graph of proportional relationship is given y =kx , answer of the following questions are as follow:
1. Based on the information ,the constant of proportionality represents the multiplicative relationship between two quantities.
2. Variable represents the constant of proportionality is k.
As given in the question,
Graph represents proportional relationship is given by:
y = kx
⇒ k = y/x
Represents the multiplicative relationship between the variables y and x.
1. Based on the information , the constant of proportionality represents the multiplicative relationship between two quantities.
'k' is the scale factor represents the constant of proportionality.
2. Variable represents the constant of proportionality is k.
Therefore, graph of proportional relationship is given y =kx , answer of the following questions are as follow:
1. Based on the information , the constant of proportionality represents the multiplicative relationship between two quantities.
2. Variable represents the constant of proportionality is k.
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Fifteen strips, 11/4" wide, are to be ripped from a sheet of plywood. If 1/8" is lost with each cut, how much of the plywood sheet is used in making the 15 strips? (Assume 15 cuts are necessary.)
The size of the plywood sheet used is;
[tex]\frac{37}{8}^{\doubleprime}[/tex]Here, we want to get the size of the part of the plywood sheet lost
From the question, we are told that 1/8 inches is lost
The size lost would be;
[tex]\frac{1}{8}\times\text{ 15 = }\frac{15}{8}[/tex]This is the size that was lost
To get the total part of the plywood used, we simply add the width of all the strips to the amount of the plywood lost
We have this as;
[tex]\frac{11}{4}\text{ + }\frac{15}{8}\text{ = }\frac{22+15}{8}\text{ = }\frac{37}{8}[/tex]When drawing a trendline, which statement is true?
A. All datasets have a trendline
B. All trendlines begin at the origin.
C. Trendlines can have a positive or negative association.
D. Trendlines have only positive associations.
Trendlines have only positive associations. Option D is correct.
Given that,
When drawing a trendline, which statement is true is to be determined.
The graph is a demonstration of curves that gives the relationship between the x and y-axis.
Here,
Trendlines are the line that explains the drastic positive change in the graph,
So Trendline has only a positive association according to the statement mentioned above.
Thus, trendlines have only positive associations. Option D is correct.
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The maintenance department at the main campus of a large state university receives daily requests to replace fluorescent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 37 and a standard deviation of 10. iS Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 37 and 67?
Answer: 49.85%
Explanation:
From the information given,
mean = 37
standard deviation = 10
The 68-95-99.7 rule states that 68% of the data fall within 1 standard deviation of the mean. 95% of the data fall within 2 standard deviations of the mean and 99.7% of the data fall within 3 standard deviations of the mean. Thus,
1 standard deviation to the left of the mean = 37 - 10 = 27
1 standard deviation to the right of the mean = 37 + 10 = 47
3 standard deviation to the left of the mean = 37 - 3(10) = 37 - 30 = 7
3 standard deviations to the right of the mean = 37 + 3(10) = 37 + 30 = 67
We can see that the percentage of lightbulb replacement requests numbering between 37 and 67 falls within 3 standard deviations to the right of the mean. This is just half of the area covered by 99.7%. Thus
The percentage of lightbulb replacement requests numbering between 37 and 67
= 99.7/2 = 49.85%
I need help with this question please. Just do question 1 please. Also this is just apart of a homework practice
Answer:
P(x) = 1.3x² + 0.1x + 2.8
Explanation:
We need to find an equation that satisfies the relationship shown in the table. So, let's replace x by 2 and then compare whether the value of p(x) is 8.2 or not
P(x) = 1.3x³ + 0.1x² + 2.8x
P(2) = 1.3(2)³ + 0.1(2)² + 2.8(2)
P(2) = 16.4
Since P(2) is 16.4 instead of 8.2, this is not a correct option
P(x) = 1.3x² + 0.2x - 2.8
P(2) = 1.3(2)² + 0.2(2) - 2.8
P(2) = 2.8
Since 2.8 and 8.2 are distinct, this is not the correct option
P(x) = 2.3x² + 0.2x + 1.8
P(x) = 2.3(2)² + 0.2(2) + 1.8
P(x) = 11.4
Since 11.4 and 8.2 are distinct, this is not the correct option
P(x) = 1.3x² + 0.1x + 2.8
P(2) = 1.3(2)² + 0.1(2) + 2.8
P(2) = 8.2
Therefore, this is the polynomial function for the data in the table.
So, the answer is P(x) = 1.3x² + 0.1x + 2.8
Analyze the equations in the graphs to find the slope of each equation the y-intercept of each equation in the solution for the system of equations equation 1: y = 50x + 122
Given:
[tex]y=50x+122\ldots\text{ (1)}[/tex][tex]y=1540-82x\ldots\text{ (2)}[/tex]The general equation is
[tex]y=mx+c[/tex]m is a slope and c is the y-intercept.
From equation (1),
[tex]\text{Slope = 50 and y intercept is 122}[/tex]From equation (2)
[tex]\text{Slope = -82 and yintercept is }1540[/tex]From equation (1) and (2)
Substitute equation (2) in (1)
[tex]1540-82x=50x+122[/tex][tex]50x+82x=1540-122[/tex][tex]132x=1418[/tex][tex]x=\frac{1418}{132}[/tex][tex]x=44[/tex]Substitute in (2)
[tex]undefined[/tex]The instructions are: Write,evaluate,graph on a Number Line the following inequalities:Six increased by twice a number is no more than 20.
• Given the description "Six increased by twice a number is no more than 20", you need to know the following:
- In this case, the word "increased" indicates an Addition.
- The word "twice" indicates a Multiplication by 2.
- "No more than" indicates that six increased by twice a number must be less than or equal to 20.
- The inequality symbol whose meaning is "Less than or equal to" is:
[tex]\leq[/tex]Knowing the information shown before, you can write the following expression to represent "Six increased by twice a number" (Let be "x" the unknown number):
[tex]6+2x[/tex]Therefore, you can write the following inequality that models the description given in the exercise:
[tex]6+2x\leq20[/tex]• Now you need to solve it:
1. Apply the Subtraction Property of Inequality by subtracting 6 from both sides of the inequality:
[tex]\begin{gathered} 6+2x-(6)\leq20-(6) \\ \\ 2x\leq14 \end{gathered}[/tex]2. Apply the Division Property of Inequality by dividing both sides of the inequality by 2:
[tex]\begin{gathered} \frac{2x}{2}\leq\frac{14}{2} \\ \\ x\leq7 \end{gathered}[/tex]• In order to graph the solution on a Number Line, you can follow these steps:
- Since the inequality symbol indicates that "x" is less than 7, it indicates that 7 is included in the solution. Therefore, you must draw a closed circle over that value.
- Draw a line from the circle to the left.
Then, you get:
Hence, the answer is:
- Inequality:
[tex]6+2x\leq20[/tex]- Solution:
[tex]x\leq7[/tex]- Number Line:
4/7 X 1/2 = in fraction
Consider the given expression,
[tex]P=\frac{4}{7}\times\frac{1}{2}[/tex]The product of fractions is obtained in the form of a fraction whose numberator is the product of numerators of fractions, and the denominator of the product is the product of denominators of the given fractions,
[tex]\begin{gathered} P=\frac{4\times1}{7\times2} \\ P=\frac{4}{14} \end{gathered}[/tex]Thus, the product of the given fractions is 4/14 .
Show your work Round to the nearest whole number if needed
Given:
Radius, r = 6
Let's find the chance of hitting the shaded area by finding the ratio.
Since the radius of the cirlce is 6, the length of one side of the square is the diameter:
s = 6 x 2 = 12
To find the ratio divide the area of the circle by area of the square. The area of the circle is the shaded area while the area of the square is the total possible area.
Thus,we have:
[tex]\text{ Area of circle = }\pi r^2=3.1416\ast6^2=3.1416\ast36=113.0976\text{ square units}[/tex][tex]\text{ Area of square = }s^2=12^2=12\ast12=144\text{ square units}[/tex][tex]\text{ Ratio=}\frac{shaded\text{ area}}{total\text{ possible area}}=\frac{area\text{ of circle}}{area\text{ of square}}=\frac{113.0976}{144}=0.7854\approx0.79[/tex][tex]\text{ Percentage ratio = 0.7854 }\ast\text{ 100=}78.54\text{ \%}[/tex]Therefore, the chance of hitting the shaded region is 78.54%
ANSWER:
78.54%
One group (A) contains 155 people. One-fifth of the people in group A will be selected to win $20 fuel cards. There is another group (B) in a nearby town that will receivethe same number of fuel cards, but there are 686 people in that group. What will be the ratio of nonwinners in group A to nonwinners in group B after the selections aremade? Express your ratio as a fraction or with a colon.
According to the information given in the exercise:
- Group A contains a total of 155 people.
- One-fifth of that people will be selected to win $20 fuel cards.
- The total number of people in Group B is 686.
Then, you can determine that the number of people that will be selected to win $20 fuel cards is:
[tex]winners_A=\frac{1}{5}(155)=31[/tex]Therefore, the number of nonwinners in Group A is:
[tex]N.winners_A=155-31=124[/tex]You know that Group B will receive the same number of fuel cards. Therefore, its number of nonwinners is:
[tex]N.winners_B=686-31=655[/tex]Knowing all this information, you can set up the following ratio of nonwinners in Group A to nonwinners in Group B after the selections are made:
[tex]\frac{124}{655}[/tex]Hence, the answer is:
[tex]\frac{124}{655}[/tex]5. What is the area of triangle ABC? (lesson 10.2)AN10 ftD 6 ftСA 15 square feetB 16 square feet© 30 square feetD 32 square feet
The answer is C, 30 square feet
Given that the two triangles are similar find the unknowns length of the side labeled in
The unknown length of the side labeled n is 10.5 units
Explanation:Given:
Two similar triangles with one unknown
To find:
the unknown length of the side labelled n
For two triangles to be similar, the ratio of their corresponding sides will equal
[tex]\begin{gathered} side\text{ with 36 corresponds to side with 27} \\ side\text{ with 14 corresponds to side with n} \\ The\text{ ratio:} \\ \frac{14}{n}\text{ = }\frac{36}{27} \end{gathered}[/tex][tex]\begin{gathered} crossmultiply: \\ 14(27)\text{ = 36\lparen n\rparen} \\ 36n\text{ = 378} \\ \\ divide\text{ both sides by n:} \\ \frac{36n}{36}\text{ = }\frac{378}{36} \\ n\text{ = 10.5} \end{gathered}[/tex]The unknown length of the side labeled n is 10.5 units
Find the complement requested angle of 10% A/ 350B/20C/170D/80
The complementary angles are angles in which the sum of them is equal to 90º
So: 90º-10º=80º
So, the complementary angle is 80º
Given: Circle PB52°РMAD =mBD =mBAC =:: 52°.: 90°:: 128°:: 142°.: 232°:: 308°
From the circle given, it can be observed that AC is the diameter of the circle and it divides the circle into two equal parts. The total angle in a semi-circle is 180°. It then follows that
[tex]arcAD+arcDC=arcAC[/tex][tex]\begin{gathered} \text{note that} \\ arcAC=180^0(\text{angle of a semicircle)} \\ arcDC=90^0(\text{given)} \end{gathered}[/tex][tex]\begin{gathered} \text{Therefore,} \\ arcAD+arcDC=arcAC \\ arcAD+90^0=180^0 \\ arcAD=180^0-90^0 \\ arcAD=90^0 \end{gathered}[/tex][tex]\begin{gathered} \text{From the circle, it can be seen that:} \\ arcBD=arcBA+arcAD \\ \text{note that } \\ arcBA=52^0(\text{given)} \\ arcAD=90^0(\text{calculated earlier)} \end{gathered}[/tex][tex]\begin{gathered} \text{Therefore,} \\ arcBD=52^0+90^0 \\ arcBD=142^0 \end{gathered}[/tex][tex]\begin{gathered} \text{From the given circle, it can be seen that} \\ arcBA+arcAD+arcDC=arc\text{BAC} \end{gathered}[/tex][tex]\begin{gathered} \text{Therefore,} \\ 52^0+90^0+90^0=\text{arcBAC} \\ 232^0=\text{arcBAC} \end{gathered}[/tex]Hence, arcAD = 90°, arc BD = 142°, and arc BAC = 232°
The graphs depict IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler IQ test).
a.find the Z score. Write that answer to the 2nd decimal place.
b. solve for x
The required Z-score with a value of 120 would be 1.33.
What is Z -score?A Z-score is defined as the fractional representation of data point to the mean using standard deviations.
The given graph depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.
As per the given information, the solution would be as
ц = 100
σ = 15
X = 120 (consider the value)
⇒ z-score = (X - ц )/σ₁
Substitute the values,
⇒ z-score = (120 - 100)/15
⇒ z-score = (20)/15
⇒ z-score = 1.33
Thus, the required Z-score with a value of 120 would be 1.33.
Learn more about the z-score here:
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1.23 × 10 to the 5th power
=
Answer:
1.23 x 10 to the 5th power is 123,000.
Step-by-step explanation:
math.
The position of an open-water swimmer is shown in the graph. The shortest route to the shoreline is one that is perpendicular to the sh Ay 10 00 6 water 4 shore |(2, 1) swimmer 19 -2 2 1 3 4 5X N -2 An equation that represents the shortest path is y=
Answer:
Explanation:
From the graph, we ca
Rewrite the equation to easily determine the velocity of an object. solve the Equation for v
In order to solve for v in the given equation, follow these steps:
1. Divide both sides of the equation by "m"
[tex]\begin{gathered} E=\frac{1}{2}mv^2 \\ \frac{E}{m}=\frac{1}{2}\frac{mv^2}{m} \\ \frac{E}{m}=\frac{1}{2}\frac{m}{m}v^2 \\ \frac{E}{m}=\frac{1}{2}v^2 \end{gathered}[/tex]2. Multiply both sides by 2
[tex]\begin{gathered} \frac{E}{m}\times2=\frac{1}{2}v^2\times2 \\ 2\frac{E}{m}=\frac{2}{2}v^2 \\ 2\frac{E}{m}=v^2 \end{gathered}[/tex]3. in order to get rid of the exponent of v, take the square root on both sides
[tex]\begin{gathered} \sqrt{2\frac{E}{m}}=\sqrt{v^2} \\ \sqrt[]{2\frac{E}{m}}=v \\ v=\sqrt[]{2\frac{E}{m}} \end{gathered}[/tex]Then, v = √(2E/m)