They are located 8 units away, so one has to be away in the left direction and the other one in the right direction
[tex]\begin{gathered} \text{ - 2 - 8 = -10 } \\ \text{ - 2 + 8 = 6} \end{gathered}[/tex]So, C and D are located at -10 and 6
Write an equation in point slope form for the line given the slope of 4,and a point on the line (1,2)
[tex]\begin{gathered} \text{ the equation of a line in slope-point form is} \\ y=mx+b,\text{ we know that m=4, and that (1,2) is on the line, so} \\ 2=4(1)+b \\ 2=4+b \\ b=2-4 \\ b=-2 \\ \\ \text{ Thus, the equation has the form} \\ y=4x-2 \\ \end{gathered}[/tex]
Answer:
[tex]y-2=4(x-1)[/tex]
Step-by-step explanation:
Pre-SolvingWe are given that a line has a slope (m) of 4, and that it contains the point (1,2).
We want to write the equation of this line in point-slope form.
Point-slope form is given as [tex]y-y_1=m(x-x_1)[/tex], where m is the slope and [tex](x_1, y_1)[/tex] is a point, hence the name.
Since we are already given the slope, we can immediately plug it into the formula.
Substitute 4 for m.
[tex]y-y_1=4(x-x_1)[/tex]
Now, let's label the values of (1,2) to avoid confusion while substituting.
[tex]x_1=1\\y_1=2\\[/tex]
Substitute these values into the formula.
[tex]y-2=4(x-1)[/tex]
Topic: point-slope form
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Liz attended class every day since she started school as a kindergarten. She said she has been in school for about 1,000 days.What numbers could be the actual number of school days if Liz rounded to the nearest ten?4 grade student
Solution.
Given that Liz attended class every day since she started school as a kindergarten and that she said she has been in school for about 1000 days (by rounding up the actual number of days to the nearest ten)
The actual number is a number that when rounded up, we would arrive at 1000.
This number falls between the numbers 995 and 1004.
Answer: Any of the numbers below could be the actual number:
995, 996, 997, 998, 999, 1000, 1001, 1002, 1003, 1004
And if you can step by step on how to do it
The radius of the cylinder is r=3 cm.
The height of the cylinder is h=7 cm.
The expression for the volume of the cylinder is,
[tex]V=\pi r^2h[/tex]Substituting the given values in the above equation,
[tex]\begin{gathered} V=\pi(3\operatorname{cm})^2(7\operatorname{cm}) \\ =\frac{22}{7}\times9cm^2\times7\operatorname{cm} \\ =198cm^3 \end{gathered}[/tex]Thus, option (C) is the correct solution.
5. The domain of f(x) = -2x + 1 is {-4, -1, 0, 2}. Find the range.
Explanation:
The function is f(x) = -2x + 1
Domain = {-4, -1, 0, 2}
Note that the domain is a set of of all the values of x ( i.e. the independent variable)
The range is a set of the corresponding value of f(x) for each value of x in the domain.
For x = -4
f(-4) = -2(-4) + 1 = 8 + 1
f(-4) = 9
f(-1) = -2(-1) + 1 = 2 + 1
f(-1) = 3
f(0) = -2(0) + 1 = 0 + 1
f(0) = 1
f(2) = -2(2) + 1 = -4 + 1
f(2) = -3
Therefore the set of all the values above which is the range will be given as:
Range = { 9, 3, 1, -3}
What is the average rate of change over the interval [1,2]Type the numerical value for your answer as a whole number, decimal, or fractionMake sure answers are completely simplified
To calculate the average rate of change over the interval [1,2] we need to identify the points in the extremes of the interval.
This points are (1,50) and (2,25).
We calculate the average rate of change as the slope:
[tex]r=\frac{y_2-y_1}{x_2-x_1}=\frac{25-50}{2-1}=-\frac{25}{1}=-25[/tex]Answer: the rate of change over [1,2] is -25.
There's a roughly linear relationship between the number of times a species of cricketwill chirp in one minute and the temperature outside. For a certain type of cricket,this relationship can be expressed using the formula I = 0.3lc + 36, where Trepresents the temperature in degrees Fahrenheit and c represents the number oftimes the cricket chirps in one minute. What is the meaning of the I'-value whenc= 148?
The function T = T(c) tells us the temperature based on the number of times c the cricket has chirped in one minute. In other words, if we plug c = 148 in the formula we get:
[tex]\begin{gathered} T(c)=0.31c+36 \\ T(148)=0.31\times(148)+36 \\ T(148)=81.88^oF \end{gathered}[/tex]That means if the cricket chirps 148 times per minute, the temperature must be 81.88 ºF.
Answer: The expected temperature in degrees Farenheit if the cricket has chirped 148 times perminute.
Zach can buy a dozen pencils for $1.89, 24 pencils for $3.60, or 36 pencils for $5.49. What is the best buy?
To know which one is the best option we have to divide the cost in the number of pencils so:
[tex]\frac{1.89}{12}=0.16[/tex]the second option is:
[tex]\frac{3.60}{24}=0.15[/tex]the thert option is:
[tex]\frac{5.49}{36}=1.16[/tex]So the best option is the second option
Determine the sum of the infinite geometric series
1/2-1/3+2/9-…
A. -1/2
B. the sum cannot be determined
C. 1/3
D. 3/10
We (B) cannot determine the sum of the given infinite geometric series (1/2-1/3+2/9-…).
What is infinite geometric series?A geometric series is one where each pair of consecutive terms' ratios is a fixed function of the summation index. The ratio is a rational function of the summation index in a more general sense creating what is known as a hypergeometric series.The result of an infinite geometric sequence is an infinite geometric series. There would be no conclusion to this series. The infinite geometric series has the general form a₁ + a₁r + a₁r² + a₁r³ +..., where r is the common ratio and a1 is the first term.So, the sum of 1/2-1/3+2/9-…
We can easily observe that the terms of the following given series are not in a series or in a particular sequence.Then, it is not possible to find the sum of this given series.Therefore, we (B) cannot determine the sum of the given infinite geometric series (1/2-1/3+2/9-…).
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evaluate the following function for f(-2) .f(x)=3x+12
Given :
[tex]f(x)=3x+12[/tex]WE need to find the value of f(-2)
So, substitute with x = -2
[tex]f(-2)=3\cdot-2+12=-6+12=6[/tex]So, the value of f(-2) = 6
Hey I need help on this math problem thank you
Answer:
From the image below we will bring out two coordinates we are going to use to calculate the rate of change of the graph
The coordinates are given below as
[tex]\begin{gathered} (x_1,y_1)\Rightarrow(2,4) \\ (x_2,y_2)\Rightarrow(-2,0) \end{gathered}[/tex]Concept:
The rate of change will be calculated using the formula below
[tex]\begin{gathered} \text{rate of change =}\frac{change\text{ in y}}{\text{change in x}} \\ \text{rate of change}=\frac{y_2-y_1}{x_2-x_1} \end{gathered}[/tex]By substituting the values, we will have
[tex]\begin{gathered} \text{rate of change}=\frac{y_2-y_1}{x_2-x_1} \\ \text{rate of change}=\frac{_{}0-4_{}}{-2_{}-2_{}} \\ \text{rate of change}=\frac{-4}{-4} \\ \text{rate of change}=1 \end{gathered}[/tex]Hence,
The rate of change = 1
Write vector h= 8i – 11j in vector component form.
Solution:
Given the vector h;
[tex]h=8i-11j[/tex]The vector in component form is;
[tex]h=<(8,-11)>[/tex]* Use the digits 0, 2, and 5 to write all of the three-digit numbers that fit each
description. You can repeat digits in a number.
multiples of 2
The 3-digit multiples of 2 using 0, 2, and 5 are:
250502520What are multiples?A multiple in science is created by multiplying any number by an integer. In other words, if b = na for some integer n, known as the multiplier, it can be said that b is a multiple of a given two numbers, a and b. This is equivalent to stating that b/a is an integer if an is not zero. In mathematics, multiples are the results of multiplying an integer by a given number. Multiples of 5 include, for instance, 10, 15, 20, 25, 30, etc. Numerous 7s include 14, 21, 28, 35, 42, 49, etc.So, 3-digit multiples of 2 using the digits 0, 2, and 5 are:
3 digits multiples of 2:
250502520Therefore, the 3-digit multiples of 2 using 0, 2, and 5 are:
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Convert each angle in radians to degrees 3π/4
we have only to change pi by 180, and solve the multiplication
[tex]\frac{3\pi}{4}\rightarrow3\cdot\frac{180}{4}=135^{\circ}[/tex]for each triangle identify a base and corresponding height use them to find the are
A)
For this tringle we can turn the figure like this:
now we have two right triangles and we can calulate the base of the first triangle with the sin law
[tex]\begin{gathered} \frac{\sin (90)}{3}=\frac{sin(a)}{2.5} \\ \sin (a)=\frac{2.5\sin (90)}{3} \\ \sin (a)=0.8 \\ a=\sin ^{-1}(0.8)=53º \end{gathered}[/tex]the angle b is going to be:
[tex]\begin{gathered} 180=90+53+b \\ b=180-90-53 \\ b=37 \end{gathered}[/tex]Now the base is going to be:
[tex]\begin{gathered} \frac{\sin(90)}{3}=\frac{\sin(37)}{\text{base}} \\ \text{base}=\frac{3\sin (37)}{\sin (90)}=1.8 \end{gathered}[/tex]and the base of the secon triangle is going to be:
[tex]\text{base}2=7.2-1.8=5.4[/tex]And the area of the triangles is going to be:
[tex]A1=\frac{base\times2.5}{2}=\frac{1.8\times2.5}{2}=2.25[/tex][tex]A2=\frac{base2\times2.5}{2}=\frac{5.4\times2.5}{2}=6.75[/tex]so in total the area is going to be:
[tex]A1+A2=2.25+6.75=9[/tex]B)
the procedure is similar, first we turn the tiangle like this:
the angle a is going to be:
[tex]\begin{gathered} \frac{\text{sin(a)}}{4.8\text{ }}=\frac{\sin (90)}{6} \\ \sin (a)=\frac{4.8\sin (90)}{6}=0.8 \\ a=\sin ^{-1}(0.8) \\ a=53º \end{gathered}[/tex]the angle b is going to be:
[tex]\begin{gathered} 180=90+53+b \\ b=180-90-53 \\ b=37º \end{gathered}[/tex]now the base is going to be:
[tex]\begin{gathered} \frac{\sin (37)}{base}=\frac{sen(90)}{4.8} \\ \text{base}=\frac{4.8\sin (37)}{\sin (90)} \\ \text{base}=2.8 \end{gathered}[/tex]and the base of the other triangle will be:
[tex]\text{base}2=5-2.8=2.2[/tex]And the area of the triangles will be:
[tex]\begin{gathered} A1=\frac{base\times4.8}{2}=\frac{2.8\times4.8}{2}=6.72 \\ A2=\frac{base2\times4.8}{2}=\frac{2.2\times4.8}{2}=5.28 \end{gathered}[/tex]And the total area will be:
[tex]A1+A2=6.72+5.28=12[/tex]Decide whether the word problem represents a linear or exponential function. Circle either linear or exponential. Then, write the function formula.
a. The given table is
Notice, the value of x increases at equal intervals of 1
Also, the value of y increases at an equal interval of 3
This means for the y values the difference between consecutive terms is 3
Also, for the x values, the difference between consecutive terms is 1
Hence, the table represents a linear function
The general form of a linear function is
[tex]y=mx+c[/tex]Where m is the slope
From the interval increase
[tex]m=\frac{\Delta y}{\Delta x}=\frac{3}{1}=3[/tex]Hence, m = 3
The equation becomes
[tex]y=3x+c[/tex]To get c, consider the values
x = 0 and y = 2
Thi implies
[tex]\begin{gathered} 2=3(0)+c \\ c=2 \end{gathered}[/tex]Hence, the equation of the linear function is
[tex]y=3x+2[/tex]b. The given table is
Following the same procedure as in (a), it can be seen that there is no constant increase in the values of y
Hence, the function is not linear
This implies that the function is exponential
The general form of an exponential function is given as
[tex]y=a\cdot b^x[/tex]Consider the values
x =0, y = 3
Substitute x = 0, y = 3 into the equation
This gives
[tex]\begin{gathered} 3=a\times b^0 \\ \Rightarrow a=3 \end{gathered}[/tex]The equation become
[tex]y=3\cdot b^x[/tex]Consider the values
x =1, y = 6
Substitute x = 1, y = 6 into the equation
This gives
[tex]\begin{gathered} 6=3\cdot b^1 \\ \Rightarrow b=\frac{6}{3}=2 \end{gathered}[/tex]Therefore the equation of the exponential function is
[tex]y=3\cdot2^x[/tex]c. The given table is
As with (b) above,
The function is exponential
Using
[tex]y=a\cdot b^x[/tex]When
x = 0, y = 10
This implies
[tex]\begin{gathered} 10=a\cdot b^0 \\ \Rightarrow a=10 \end{gathered}[/tex]The equation becomes
[tex]y=10\cdot b^x[/tex]Also, when
x = 1, y =5
The equation becomes
[tex]\begin{gathered} 5=10\cdot b^1 \\ \Rightarrow b=\frac{5}{10} \\ b=\frac{1}{2} \end{gathered}[/tex]Therefore, the equation of the exponential function is
[tex]y=10\cdot(\frac{1}{2})^x[/tex]Aunt Eloise’s house is always 20°C. She has just made a fresh cup of tea (tea is made with boiling water and water boils at 100°C) five minutes after she made the tea her mad scientist nephew came in, stuck a thermometer in the cup and announced that the tea was now only 70°C. She had gotten involved with her book and forgot to have even a sip of her tea. Now she won’t drink it because it isn’t piping hot anymore.Write and equation that models this problem and use it to predict the temperature of the tea 20 minutes after it was taken off the stove.
Given:
a.) She has just made a fresh cup of tea (tea is made with boiling water and water boils at 100°C)
b.) Five minutes after she made the tea her mad scientist nephew came in, stuck a thermometer in the cup, and announced that the tea was now only 70°C.
c.)
Which transformations can be used to carry ABCD onto itself? The point ofrotation is (3,2). Check all that apply.у5DС321АВ012345
The point of rotation is (3, 2). We would subtract this origin from each vertex. Let us consider vertex D(1, 3). Subtracting, we have (1 - 3, 3 - 2) = (- 2, 1).
if we rotate it 180 degrees, we have (- - 2, - 1) = (2, - 1). If we add the vertex again, it becomes (- 2 + 3, 1 + 2) = (1, 3). If we reflect it over the line, y = 2, we have (3, 2)
The correct options are
B. Rotate 180 degrees
D) reflection over the line, y = 2
what percentage of students scored before 70-90 points on the exam? Round your answer to the nearest tenth of a percent?
We want to find the percentage of students that scored between 70-90 points on the examn. Also, we know that there are a total of 71 students, so we have to count the number of students who got between 70-90 points.
We see them represented on the histogram as the two largest bars, and we obtain:
[tex]\begin{gathered} 21\text{ students scored between 70-80 points} \\ 22\text{ students scored between 80-90 points} \end{gathered}[/tex]So, the total of students that scored between 70-90 points is 21+22=43.
For finding the percentage, we make the quotient between the number of students with a score of 70-90 and the total of the students in the class.
[tex]=\frac{43}{71}\cdot100=60.6[/tex]This means that approximately a 60.6% of the class students scored between 70 and 90 points.
Questlon 5 Refer to the figure. HJ I JE. HII IE. HJ HI J H E Complete the explanation to show triangle EJH is congruent to triangle EIH. The two triangles given are _____triangles. The leg and hypotenuse of triangle EJH are congrue hypotenuse of triangle EIH. By the ______ Theorem the third side ma triangles are congruent by the____ Triangle Congruence Theorem.choice 1.acute,obtuse or right angleschoice 2.corresponding parts of congruent triangles, pythagorean,or side-angle-side triangle congruence.choice 3. side-side-side, side-angle-side,or angle-side-angle
In the given figure, we have two triangles △EJH and △EIH
We are given the following information
[tex]\begin{gathered} \bar{HJ}\perp\bar{JE} \\ \bar{HI}\perp\bar{IE} \\ \bar{HJ}\cong\bar{HI} \end{gathered}[/tex]This means that these two triangles are "Right Triangles"
Therefore.
Choice 1 = right angles
When the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of a second right triangle, then the triangles are congruent.
Therefore,
Choice 2 = side-angle-side triangle congruence
Choice 3 = side-angle-side
Answer: choice 1 -Right Angle
Choice 2 -Pythagorean
Choice 3- Side-Side-Side
Step-by-step explanation:other guy is completely wrong lol
solving rational equations 1[tex] \frac{9x}{x - 2} = 6[/tex]2[tex] \frac{7}{x + 2} + \frac{5}{x - 2} = \frac{10x - 2}{x {}^2{ - 4} } [/tex]3[tex] \frac{ x - 1}{2x - 4} = \frac{2x - 2}{3x} [/tex]step by step instructions please thank you
the given expression is.,
[tex]\frac{9x}{x-2}=6[/tex][tex]\begin{gathered} 9x=6x-12 \\ 9x-6x=-12 \\ 3x=-12 \\ x=-\frac{12}{3} \\ x=-4 \end{gathered}[/tex]also, the given expression is,
[tex]\frac{7}{x+2}+\frac{5}{x-2}=\frac{10x-2}{x^2-4}[/tex][tex]\begin{gathered} \frac{7x-14+5x+10}{x^2-4}=\frac{10x-2}{x^2-4} \\ 12x-4=10x-2 \\ 12x-10x=4-2 \\ 2x=2 \\ x=\frac{2}{2}=1 \\ x=1 \end{gathered}[/tex]Classify the triangle formed by the side lengths as right, acute, or obtuse. 11 13 23
The triangle of side lengths 11, 13 and 23 is obtuse triangle.
Since none of the sides are equal.
If p(x) is a polynomial function where p(x) = 3(x + 1)(x - 2)(2x-5)a. What are the x-intercepts of the graph of p(x)?b. What is the end behavior (as x→ ∞, f(x)→?? and as x→ -∞, f (x)→ ??) of p(x))?c. Find an equation for a polynomial q(x) that has x-intercepts at -2, 3⁄4, and 7.
Hello there. To solve this question, we have to remember some properties about polynomial functions.
Given the polynomial function
[tex]p(x)=3(x+1)(x-2)(2x-5)[/tex]We want to determine:
a) What are the x-intercepts of the graph of p(x)?
For this, we have to determine the roots of the polynomial function p(x). In this case, we have to determine for which values of x we have
[tex]p(x)=0[/tex]Since p(x) is written in canonical form, we find that
[tex]p(x)=3(x+1)(x-2)(2x-5)=0[/tex]A product is equal to zero if at least one of its factors is equal to zero, hence
[tex]x+1=0\text{ or }x-2=0\text{ or }2x-5=0[/tex]Solving the equations, we find that
[tex]x=-1\text{ or }x=2\text{ or }x=\dfrac{5}{2}[/tex]Are the solutions of the polynomial equation and therefore the x-intercepts of p(x).
b) What is the end-behavior of p(x) as x goes to +∞ or x goes to -∞?
For this, we have to take the limit of the function.
In general, for polynomial functions, those limits are either equal to ∞ or -∞, depending on the degree of the polynomial and the leading coefficient.
For example, a second degree polynomial function with positive leading coefficient is a parabola concave up and both limits for the function as x goes to ∞ or x goes to -∞ is equal to ∞.
On the other hand, an odd degree function usually has an odd number of factors (the number of x-intercepts in the complex plane) hence the limits might be different.
In this case, we have a third degree polynomial equation and we find that, as the leading coefficient is positive and all the other factors are monoic, that
[tex]\begin{gathered} \lim_{x\to\infty}p(x)=\infty \\ \\ \lim_{x\to-\infty}p(x)=-\infty \end{gathered}[/tex]That is, it gets larger and larger when x is increasing arbitrarily, while it get smaller and smaller as x is decreasing.
c) To find the equation for a polynomial q(x) that has x-intercepts at -2, 3/4 and 7.
The canonical form of a polynomial of degree n with x-intercepts at x1, x2, ..., xn and leading coefficient equals a is written as
[tex]f(x)=a\cdot(x-x_1)(x-x_2)\cdots(x-x_n)[/tex]So in this case, there are infinitely many polynomials satisfying this condition. Choosing a = 1, we find that q(x) is equal to
[tex]\begin{gathered} q(x)=(x-(-2))\cdot\left(x-\dfrac{3}{4}\right)\cdot(x-7) \\ \\ \boxed{q(x)=(x+2)\cdot\left(x-\dfrac{3}{4}\right)\cdot(x-7)} \end{gathered}[/tex]These are the answers to this question.
help pleasesjjsnsbsbbbsbs
Student asking the same question for third time in less than ten minutes. Can't help him or her out with additional information to complete the exercise.
Closing the session!
The original price of a sweater is $48. The sale price is 80%. What is the sale price of the sweater?
Letters a, b, c, and d are angle measures. Which should equal 105° to prove that fll g? Фа Ob n b 75° 0 d g f Mark this and return Save and Exit Next Submit
in the given figure,
the sum of exterior angle 75 and d will be 180
we have 75 + d = 180
d = 180 - 75
d = 105 degrees,
thus, the correct answer is option D
The Oldest rocks on Earth are about 4 x 10^9 years old. For which of these ages could this be an approximation?
A. 3,862,100,000 years
B. 3.849999999x10^9 years
C. 0.000000004 years
D.4,149,000,000 years
E.3.45x10^9 years
what is the domain and range of arccosine?Thanks!
Solution:
What is the domain and range of arccosine?
The function is
[tex]f(x)=\cos^{-1}(x)[/tex]The graph of the function is shown below
From the graph above;
The domain of the function is
[tex]-1\leq x\leq1[/tex]The range of the function is
[tex]0\leq y\leq\pi[/tex]Write the coordinates of the vertices after a reflection over the line y=-x
The coordinates of the vertices after a reflection over the line y=-x are (y, -x)
How to determine the coordinates?From the question, the transformation rule is given as
Reflection over the line y=-x
There are four types of transformation,
These transformations are
DilationRotationReflectionTranslationEach of these transformations have their rule, and they are represented as
Reflection: reflection across linesDilation: k(x, y)Rotation: rotation by anglesTranslation: (x + h, y + k)So, we have
Reflection over the line y=-x
When represented as a coordinate, the coordinate is
(x, y) = (y, -x)
Hence, the coordinates are (y, -x)
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Draw the graph of the line that is parallel to Y -3 = 1/3(x+2) and goes through the point (1, 7)
Explanation:
We are required to draw the graph of the line that is parallel to y-3=1/3(x+2) and goes through the point (1, 7).
Given the equation of the line:
[tex]y-3=\frac{1}{3}(x+2)[/tex]Compare the equation with the slope-point form of a line:
[tex]$$y-y_1=m(x-x_1)$$[/tex]• The slope of the line, m=1/3
,• In addition, the line goes through the point (1,7)
Substitute these values into the point-slope form given above:
[tex]y-7=\frac{1}{3}(x-1)[/tex]Finally, graph the line by looking for another point in addition to point (1,7):
When x=-2
[tex]\begin{gathered} y-7=\frac{1}{3}(x-1) \\ y-7=\frac{1}{3}(-2-1) \\ y-7=\frac{1}{3}(-3) \\ y-7=-1 \\ y=-1+7 \\ y=6 \\ \implies(-2,6) \end{gathered}[/tex]Join the points (1, 7) and (-2, 6) to plot the line.
Answer:
The graph showing the two points is attached below:
Note:
For comparison purposes and to show that the two lines are parallel, the other graph is added below:
5x^2+2x-3 and x+4 how do I find the area
You have a rectangle with the followinf expressions for its sides:
w: width = 5x² + 2x - 3
h: height = x + 4
In order to calculate the area of the rectangle, you use the following formula for the area:
A = w*h
By replacing w and h by the given algebraic expressions you have:
A = (5x² + 2x - 3)(x + 4) use distribution property
A = 5x²(x) + 5x²(4) + 2x(x) + 2x(4) - 3(x) - 3(4) simplify
A = 5x³ + 20x² + 2x² + 8x - 3x - 12 simplify similar terms
A = 5x³ + 22x² + 5x - 12
Hence, the total area of the figure is 5x³ + 22x² + 5x - 12