To find:
We need to find two points on the linear equation y-1=0 and to plot those points on graph.
Step by step solution:
We know that:
General coordinate of any two points on line y = 1:
= (x, 1)
So let us assume any two random points on the line:
= (1,1) and (2,1)
We will now mark them on the graph:
Gor trapezoid HJKL, T and S are midpoint of the legs. If HJ = 14 and LK = 42, find TS.
First, we are going to divide the figure and named new points X and Y as:
Now, we know that TS is the sum of TX and XS.
TS = TX + XS
Adittionally, TX has the same length of HJ, so:
TX = HJ = 14
Now, we want to know the length of YK, and we can calculate it using the following equation:
LK = LY + YK LY is also equal to HJ, so LY = 14
42 = 14 + YK
42 - 14 = YK
28 = YK
Finally, since T and S are midpoints, the length of XS is the half of the length of YK. It means that XS is:
XS = YK/2
XS = 28/2
XS = 14
Therefore, TS is equal to:
TS = TX + XS
TS = 14 + 14
TS = 28
Answer: TS = 28
How does basic algebra come to play in everyday life? Explain (or give examples) in at least two sentences
Explanation
1) Algebra can be used while cooking to estimate the amount of ingredients by solving some easy algebraic expressions of the head.
e.g 2 tea spoons of pepper out of a 1kg pack might be the right amount to spice a soup.
2) For example, a plumber may do some quick calculations to determine the number of pipes required for a house
e.g 5 pipes in the bathroom, two pipes in the toilet, three in the kitchen gives 10 pipes altogether.
how to find two consecutive whole numbers that square root 40 lies between
First, we need to identify the square root of the fisrt squared numbers:
[tex]\begin{gathered} \sqrt{1}=\text{ 1} \\ \sqrt{4}=2 \\ \sqrt{9}=3 \\ \sqrt{16}=4 \\ \sqrt{25}=5 \\ \sqrt{36}=6 \\ \sqrt{49}=7 \\ \sqrt{64}=8 \end{gathered}[/tex]Since 40 is a number between 36 and 49, we can say that the square root of 40 is between 6 and 7. So:
[tex]6<\sqrt{40\text{ }}<7[/tex]Victoria and her children went into a grocery store and she bought $9 worth of applesand bananas. Each apple costs $1.50 and each banana costs $0.50. She bought a totalof 8 apples and bananas altogether. Determine the number of apples, x, and thenumber of bananas, y, that Victoria bought.Victoria boughtapples andbananas.
We will determine the solution as follows:
*First: From the text, we have the following expressions:
[tex]x+y=8[/tex]&
[tex]1.50x+0.5y=9[/tex]Here x represents apples and y represents bananas.
*Second: From the first expression, we solve for either x or y, that is [I will solve for ]:
[tex]x+y=8\Rightarrow x=8-y[/tex]*Third: Now, using the value for x, we replace in the second expression and solve for y, that is:
[tex]1.50x+0.5y=9\Rightarrow1.50(8-y)+0.5y=9[/tex][tex]\Rightarrow12-1.50y+0.5y=9\Rightarrow-y=-3[/tex][tex]\Rightarrow y=3[/tex]*Fourth: We replace the found value of y on the first expression and solve for x:
[tex]x+y=8\Rightarrow x+3=8[/tex][tex]\Rightarrow x=5[/tex]So, the number of apples was 5 and the number of bananas was 3.
The Hornet's soccer team scored 5 goals in their last match.The other team, the Panthers, won by 3 goals. Which integerrepresents the number of goals that the Panthers won by?
The match was Hornet's vs Panthers
Hornets's scored 5 goals
Panthers won by 3 goals, this means that the panters scored 3 more goals than the Hornets.
That would be +3 goals.
Graph of this line using intercepts. I need some help some assistance would be nice
Explanation:
The equation of the line is given below as
[tex]2x+3y=18[/tex]Step 1:
To determine the x-intercepts, we will put y=0 and solve for x
[tex]\begin{gathered} 2x+3y=18 \\ 2x+3(0)=18 \\ 2x+0=18 \\ 2x=18 \\ \frac{2x}{2}=\frac{18}{2} \\ x=9 \\ x-intercept=(9,0) \end{gathered}[/tex]Step 2:
To determine the y-intercept, we will put x=0 and solve for y
[tex]\begin{gathered} 2x+3y=18 \\ 2(0)+3y=18 \\ 0+3y=18 \\ \frac{3y}{3}=\frac{18}{3} \\ y=6 \\ y-intercept=(0,6) \end{gathered}[/tex]Hence,
The graph using the intercepts will be given below as
Adina sets up a taste test of 3 different waters: tap, bottled in glass, and bottled in plastic. She puts these waters in identical cups and has a friend taste them one by one. The friend then tries to identify which water was in each cup. Assume that Adina's friend can't taste any difference and is randomly guessing. What is the probability that Adina's friend correctly identifies each of the 3 cups of water
Given
3 different waters: tap, bottled in glass, and bottled in plastic.
Find
probability that Adina's friend correctly identifies each of the 3 cups of water
Explanation
As we have given three different waters : tap , bottled in glass and bottled in plastic.
number of ways in which the person can make guesses about the 3 cups of water =
[tex]\begin{gathered} ^3P_3 \\ \frac{3!}{0!} \\ 6 \end{gathered}[/tex]number of ways in which person identifies correctly the 3 cups of water = 1
so , probability that Adina's friend correctly identifies each of the 3 cups of water =
[tex]P\text{ = }\frac{number\text{ of ways in which person identifies correctly the 3 cups of water}}{number\text{ of ways in which the person can make guesses about the 3 cups of water }}[/tex]so , P = 1/6
Final Answer
Therefore , the probability that adina's friend correctly identifies each of the cup of water = 1/6
Find the center, vertices, foci, endpoints of the latera recta and equations of the directrices. Then sketch the graph of the ellipse.
The given equation of ellipse is,
[tex]\frac{(x-2)^2}{16}+\frac{y^2}{4}=1\text{ ---(1)}[/tex]The above equation can be rewritten as,
[tex]\frac{(x-2)^2}{4^2}+\frac{y^2}{2^2}=1\text{ ----(2)}[/tex]The above equation is similar to the standard form of the ellipse with center (h, k) and major axis parallel to x axis given by,
[tex]\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1\text{ ----(3)}[/tex]where a>b.
Comparing equations (2) and (3), h=2, k=0, a=4 and b= 2.
Hence, the center of the ellipse is (h, k)=(2, 0).
The coordinates of the vertices are given by,
[tex]\begin{gathered} (h+a,\text{ k)=(2+}4,\text{ }0)=(6,\text{ 0)} \\ (h-a,\text{ k)=(2-}4,\text{ }0)=(-2,\text{ 0)} \end{gathered}[/tex]Hence, the coordinates of the vertices are (6, 0) and (-2,0).
The coordinates of the co-vertices are given by,
[tex]\begin{gathered} (h,\text{ k+}b)=(2,\text{ }0+2)=(2,\text{ 2)} \\ (h,\text{ k-}b)=(2,\text{ }0-2)=(2,\text{ -2)} \end{gathered}[/tex]Hence, the coordinates of the co-vertices are (2, 2) and (2, -2).
The coordinates of the foci are (h±c, k).
[tex]\begin{gathered} c^2=a^2-b^2 \\ c^2=4^2-2^2 \\ c^2=16-4 \\ c^2=12 \\ c=2\sqrt[]{3} \end{gathered}[/tex]Using the value of c, the coordinates of the foci are,
[tex]\begin{gathered} \mleft(h+c,k\mright)=(2+2\sqrt[]{3},\text{ 0)} \\ (h-c,k)=(2-2\sqrt[]{3},\text{ 0)} \end{gathered}[/tex]Therefore, the coordinates of the foci are,
[tex](2+2\sqrt[]{3},\text{ 0) and }(2-2\sqrt[]{3},\text{ 0)}[/tex]The endpoints of the latus rectum is,
[tex]\begin{gathered} (h+c,\text{ k}+\frac{b^2}{a})=(2+2\sqrt[]{3},\text{ 0+}\frac{2^2}{4^{}}) \\ =(2+2\sqrt[]{3},\text{ 1)}^{} \\ (h-c,\text{ k}+\frac{b^2}{a})=2-2\sqrt[]{3},\text{ 0+}\frac{2^2}{4^{}}) \\ =(2-2\sqrt[]{3},\text{ 1}^{}) \\ (h+c,\text{ k-}\frac{b^2}{a})=(2+2\sqrt[]{3},\text{ 0-}\frac{2^2}{4^{}}) \\ =(2+2\sqrt[]{3},\text{ -1}^{}) \\ (h-c,\text{ k-}\frac{b^2}{a})=(2-2\sqrt[]{3},\text{ 0-}\frac{2^2}{4^{}}) \\ =(2-2\sqrt[]{3},\text{ -1}^{}) \end{gathered}[/tex]Therefore, the coordinates of the end points of the latus recta is,
[tex](2+2\sqrt[]{3},\text{ 1)},\text{ }(2-2\sqrt[]{3},\text{ 1}^{}),\text{ }(2+2\sqrt[]{3},\text{ -1}^{})\text{ and }(2-2\sqrt[]{3},\text{ -1}^{})[/tex]Now, the equations of the directrices is,
[tex]\begin{gathered} x=h\pm\frac{a}{e} \\ x=\pm\frac{a}{\sqrt[]{1-\frac{b^2}{a^2}}} \\ x=2\pm\frac{4}{\sqrt[]{1-\frac{2^2}{4^2}}} \\ x=2\pm\frac{4}{\sqrt[]{1-\frac{1^{}}{4^{}}}} \\ x=2\pm\frac{4}{\sqrt[]{\frac{3}{4}^{}}} \\ x=2\pm4\sqrt[]{\frac{4}{3}} \end{gathered}[/tex]Here, e is the eccentricity of the ellipse.
Therefore, the directrices of the ellipse is
[tex]x=2\pm4\sqrt[]{\frac{4}{3}}[/tex]Now, the graph of the ellipse is given by,
Given the recursive formula for an arithmetic sequence,An = an-1 - Tt, where the first term of the sequence is 7. Which of the following could be explicitformulas for the sequence? Select all that apply.
From the recursive formula:
[tex]a_n=a_{n-1}-\pi[/tex]we notice that the common difference of the sequence is -pi. Now we know that the first term is 7, then the explicit formula is:
[tex]a_n=7-\pi(n-1)[/tex]when
[tex]n>0[/tex]We can relabel this sequence if we assume we start at zero, in this case the sequence will be:
[tex]a_n=7-\pi n[/tex]when:
[tex]n\ge0[/tex]Find the equation of the line, in slope-intercept form, that passes through the points (-2, -4) and (2,8).A) y = 1/3x + 22/3B) y = 3x + 14C) y = 3x + 2 D) y = - 3x + 14
The equation of a line in the slope intercept form is expressed as
y = mx + c
where
m = slope
c = y intercept
The formula for calculating slope is expressed as
m = (y2 - y1)/(x2 - x1)
where
x1 and y1 are the x and y coordinates of the initial point
x2 and y2 are the x and y coordinates of the final point
From the information given, the initial point is (- 2, - 4) and final point is (2, 8)
Thus,
x1 = - 2, y1 = - 4
x2 = 2, y2 = 8
By substituting these values into the slope formula,
m = (8 - - 4)/(2 - - 2) = (8 + 4)/(2 + 2) = 12/4 = 3
We would find the y intercept, c by substituting m = 3, x = - 2 and y = - 4 into the slope intercept equation. We have
- 4 = 3 * - 2 + c
- 4 = - 6 + c
Adding 6 to both sides of the equation,
- 4 + 6 = - 6 + 6 + c
c = 2
By substituting m = 3 and c = 2 into the slope intercept equation, the equation of the line is
C) y = 3x + 2
Julie can run 3 laps in 9 minutes. At this rate, how many laps can she run in 24 minutes?
Answer:
Julie can run 12 laps
Step-by-step explanation:
9 min = 3 laps
9 x 2 = 18 = 6 laps
9 cant fit into 24 again
24 - 18 = 6
6 + 6 = 12 laps
a road is 4/7 of a mile long. a crew needs to replace 4/5 of the road. how long is the section that needs to be repaired
To solve this problem we need to find the fraction of a fraction, for that we just have to multiply them. This is done below:
[tex]\frac{4}{7}\cdot\frac{4}{5}=\frac{16}{35}\text{ of a mile}[/tex]The section is 16/35 of a mile long.
2×+22=2(x+11)whats the property
Distributive property
In this property, multiplying the sum of two or more terms in that add up in a bracket by a number outside the bracket will be equal to multiplying each term in the bracket individually and then followed by sum of the product. In this question:
2x + 22 = 2(x + 11 ) in that when you perform product on the right side of the equation, the result is the same i.e 2x + 2*11 = 2x + 22
What is 1/3 of the sum of 45 and a number is 16 Translated to algebraic equation
the statement given 1/3 of the sum of 45 and a number is 16
[tex]\frac{1}{3}(45+x)=16[/tex]In order to find the value of x, w
For circle H, JN = x, NK = 8, LN = 4, and NM = 20.Solve for x.
Solution
Consider the illustration below
Using the idea of the illustration above,
[tex]JN\text{ x NK = LN x NM}[/tex][tex]\begin{gathered} x\text{ x 8 = 4 x 20} \\ 8x=80 \\ x=\frac{80}{8} \\ x=10 \end{gathered}[/tex]The answer is 10
15. [-/1 Points]DETAILSCURRENMEDMATH11 2.9.027.Divide the fraction. Express your answer to the nearest tenth. A calculator may be used.180,000120,000eBook16. [-/1 Points]DETAILSCURRENMEDMATH11 2.3.028.Divide the fraction. Express your answer to the nearest tenth. A calculator may be used.0.110.08eBook
You have the following fraction:
180000/120000
First of all you cancel zeros:
180000/120000 = 18/12
next, you can simplify
18/12 = 9/6 = 3/2
finally 3/2 is:
3/2 = 1.5
Hence: 180000/120000 = 1.5
Furthermore, for the following fraction:
0.11/0.08
Here, you can use a calculator. The result is:
0.11/0.08 = 1.375
that is approximately
1.375 ≈ 1.4
For other fractions:
350/10,000 = 35/1,000 = 0.035
which is approximately
0.035 ≈ 0.04
6.01/7.2 = 0.834 ≈ 0.83
Solve the equation. f(x)=g(x) by graphing. f(x) = l x +5 l g(x) = 2x + 2 Select all possible solutions: No Solutions x=3 x=0 X=-1
As you can observe in the graph below, the given functions intercept at one point.
Hence, there is a unique solution and it's x = 3.6. Find the domain and range of V(x) in this context.7. Think of V(x) as a general function without the constraint of modeling the volume of a box. What would be the domain and range of V(x)?8. Use correct notation to describe the end behavior of V(x) as a function without context.
We have , that measure of the side of the square is x
Therefore
l=26-2x
w=20-2x
h=x
Therefore the Volume function is
[tex]V=(26-2x)(20-2x)x[/tex]Then we simplify
[tex]V(x)=4x^3-92x^2+520x[/tex]6.In the context of obtaining a Volume we can't have negative numbers for x and for the function by observing the graph
Domain
[tex]0\le x\le10[/tex]Therefore for the range
[tex]0\: 7.Because we have a polynomial
the domain without the constrain
[tex]-\infty\: the range without the constrain[tex]-\infty\: 8.Since the leading term of the polynomial is 4 x^{3}, the degree is 3, i.e. odd, and the leading coefficient is 4, i.e. positive. This means
[tex]\begin{gathered} x\to-\infty,\text{ }f(x)\to-\infty \\ x\to\infty,f(x)\to\infty \end{gathered}[/tex]Last year, Trey opened an investment account with $8800. At the end of the year, the amount in the account had decreased by 6.5%. How much is this decrease in dollars? How much money was in his account at the end of last year?Decrease in amount:$Year-end amount:$
ANSWER
[tex]\begin{gathered} decrease=572 \\ Year-end\text{ amount=8228} \end{gathered}[/tex]EXPLANATION
Initial amount is $8800
percentage decrease is 6.5%
Decrease amount (in dollars );
[tex]\begin{gathered} \frac{6.5}{100}\times8800 \\ =6.5\times88 \\ =572 \end{gathered}[/tex]The amount of money in the account at the end of last year= Initial amount - decrease
[tex]\begin{gathered} A=8800-572 \\ =8228 \end{gathered}[/tex]Decrease in amount: $572
Year-end amount: $8228
Michael withdraws $40 from his checking account each day how long will it take him to withdraw $680
Solution
- The amount Michael withdraws every day is $40.
- The number of days it takes to withdraw $680 is given by:
[tex]\frac{\text{Total Amount Withdrawn}}{\text{Amount Withdrawn per day}}[/tex]- Using the formula above, we have:
[tex]\frac{\text{Total Amount Withdrawn}}{\text{Amount Withdrawn per day}}=\frac{680}{40}=17\text{ days}[/tex]Final Answer
The answer is 17 days
The volume of a rectangular prism is 2 x cubed + 9 x squared minus 8 x minus 36 with height x + 2. Using synthetic division, what is the area of the base?
The base area of the prism is 2x² + 5x - 18
How to determine the area of the base?From the question, the given parameters are
Volume = 2 x cubed + 9 x squared minus 8 x minus 36
Height = x + 2
Rewrite properly as
Volume = 2x³ + 9x² - 8x - 36
Height = x + 2
The base area is calculated as
Base area = Volume/Height
Using the synthetic division, we have
Set the divisor to 0
x + 2 = 0
This gives
x = -2
So, we have the representation to be
-2 | 2 9 - 8 - 36
Write out 2
So, we have
-2 | 2 9 - 8 - 36
2
Multiply 2 and -2
This gives
-2 | 2 9 - 8 - 36
-4
2
So, we have
-2 | 2 9 - 8 - 36
-4
2 5
Repeat the process
So, we have
-2 | 2 9 - 8 - 36
-4 -10
2 5 -18
Repeat the process
So, we have
-2 | 2 9 - 8 - 36
-4 -10 36
2 5 -18 0
This means that
Base area = 2x² + 5x - 18
Read more about synthetic division at
https://brainly.com/question/24629353
#SPJ1
Last year, Bob had $10,000 to invest. He invested some of it in an account that paid 10% simple interest per year, and he invested the rest in an account that paid 8% simple interest per year. After one year, he received a total of $820 in interest. How much did he invest in each account?
Given:
The total amount is P = $10,000.
The rate of interest is r(1) = 10% 0.10.
The other rate of interest is r(2) = 8%=0.08.
The number of years for both accounts is n = 1 year.
The total interest earned is A = $820.
The objective is to find the amount invested in each account.
Explanation:
Consider the amount invested for r(1) as P(1), and the interest earned as A(1).
The equation for the amount obtained for r(1) can be calculated as,
[tex]\begin{gathered} A_1=P_1\times n\times r_1 \\ A_1=P_1\times1\times0.1 \\ A_1=0.1P_1\text{ . . . . .(1)} \end{gathered}[/tex]Consider the amount invested for r(2) as P(2), and the interest earned as A(2).
The equation for the amount obtained for r(2) can be calculated as,
[tex]\begin{gathered} A_2=P_2\times n\times r_2 \\ A_2=P_2\times1\times0.08 \\ A_2=0.08P_2\text{ . . . . . (2)} \end{gathered}[/tex]Since, it is given that the total interest earned is A=$820. Then, it can be represented as,
[tex]A=A_1+A_2\text{ . . . . . (3)}[/tex]On plugging the obtained values in equation (3),
[tex]820=0.1P_1+0.08P_2\text{ . . . . .(4)}[/tex]Also, it is given that the total amount is P = $10,000. Then, it can be represented as,
[tex]\begin{gathered} P=P_1+P_2 \\ 10000=P_1+P_2 \\ P_1=10000-P_2\text{ . }\ldots\ldots.\text{. .(3)} \end{gathered}[/tex]Substitute the equation (3) in equation (4).
[tex]undefined[/tex]Jelani filled an aquarium with blocks that were each one cubic foot in size. He filled the bottom layer of the aquarium with 21 blocks. He then stacked three more blocks on top of the bottom layer. The partially filled aquarium is shown below. What is the total volume, in cubic feet, of the aquarium?
Answer:
The total volume of the aquarium is;
[tex]84\text{ }ft^3[/tex]Explanation:
Given the figure in the attached image.
The bottom of the aquarium was covered with 21 blocks with 1 cubic foot each.
Each face of the cubic blocks will have a surface area of 1 square foot each.
So, the surface area of the base of the aquarium will be;
[tex]\begin{gathered} A=21\times1ft^2 \\ A=21\text{ }ft^2 \end{gathered}[/tex]Recall that volume equals base area multiply by the height of the aquarium;
[tex]V=A\times h[/tex]From the figure, the height of the aquarium requires 4 blocks, which makes the height 4 ft;
[tex]h=4ft[/tex]So, we can now substitute the values of the height and the base area to calculate the total volume of the aquarium;
[tex]\begin{gathered} V=A\times h \\ V=21ft^2\times4ft \\ V=84\text{ }ft^3 \end{gathered}[/tex]Therefore, the total volume of the aquarium is;
[tex]84\text{ }ft^3[/tex]Which of the following are greater than 1/443% 5/90.151/121.4
To solve this, we will need to convert all values to decimal.
First convert 1/4 to decimal:
[tex]\frac{1}{4}\text{ = 0.25}[/tex]Now convert 43% to decimal:
[tex]43\text{percent = }\frac{43}{100}\text{ = 0.43}[/tex]Simplify 5/9:
[tex]\frac{5}{9}\text{ = }0.56[/tex]0.15 is already a decimal value.
Simplify 1/12:
[tex]\frac{1}{12}\text{ = 0.083}[/tex]1.4 is already a decimal.
After simplifying, we have the following values:
0.25
0.43
0.56
0.15
0.083
1.4
We can see the values greater than 0.25 are:
0
Can you please answer this question for me. I don’t want full explanation I just want the answers
we have the fractions
1/4 and 3/4
Remember that
If the denominators are the same, then the fraction with the greater numerator is the greater fraction
3/4 > 1/4
use the number line
Divide number 1 into 4 parts
The sum of two numbers is 164. The second number is 24 less than three times the first number. Find the numbers.
13. (08.05 MC)What is the shape of the cross section taken perpendicular to the base of a cylinder? (1 point)CircleRectangleSquareTriangle
ANSWER
Rectangle
EXPLANATION
If we take a cross-section perpendicular to the base of a cylinder,
We get a rectangle.
The base of the cylinder is circular and the
Answer: Rectangle
Step-by-step explanation:
I got it right on the test
help me please. using the axis of symmetry find the vertex for the follow quadratic function. f (x)=3x^2-6x+8
Answer:
[tex]P(1,5)[/tex]
Explanation: Axis of symmetry is a vertical line that makes function symmetrical along either side:
In case of parabla function or:
[tex]y(x)=3x^2-6x+8[/tex]We get axial symmetry where the first derivate is zero, and in fact, that is the x value for vertex:
Therefore:
[tex]\begin{gathered} f^{\prime}(x)=(3x^2-6x+8)^{\prime}=6x-6=0 \\ \therefore\rightarrow \\ x=\frac{6}{6}=1 \end{gathered}[/tex]And the corresponding y-value is:
[tex]f(1)=3(1)^2-6(1)+8=5[/tex]Therefore vertex is at the point:
[tex]P(1,5)[/tex]Can someone help me with this geometry question? I will provide more information.
So you are given a triangle ABC and you need to build another one DEF that meets the following:
[tex]\begin{gathered} AB=DE \\ m\angle E=90^{\circ} \\ EF=BC \end{gathered}[/tex]First of all we should find the lengths of sides AB and BC. For this purpose we can use the coordinates of points A, B and C. The length of AB is the distance between A and B and the length of BC is the distance between B and C. The distance between two generic points (a,b) and (c,d) is given by:
[tex]\sqrt[]{(a-c)^2+(b-d)^2}[/tex]Then the length of AB is:
[tex]AB=\sqrt[]{(1-1)^2+(6-1)^2}=\sqrt[]{0+5^2}=5[/tex]And that of BC is:
[tex]BC=\sqrt[]{(1-5)^2+(1-1)^2}=\sqrt[]{4^2}=4[/tex]Then the triangle DEF must meet these three conditions:
[tex]\begin{gathered} DE=5 \\ EF=4 \\ m\angle E=90^{\circ} \end{gathered}[/tex]Since there is no rules about its position we can draw it anywhere. For example you can choose E=(-4,1). Then if D=(-4,6) we have that the length of DE is 5:
[tex]DE=\sqrt[]{(-4-(-4))^2+(6-1)^2}=\sqrt[]{0+5^2}=5[/tex]And if we take F=(0,1) we get EF=4:
[tex]EF=\sqrt[]{(-4-0)^2+(1-1)^2}=\sqrt[]{16}=4[/tex]Then a possibility for triangle DEF is:
As you can see it also meets the condition that the measure of E is 90°. And that would be part A.
In part B we have to use the pythagorean theorem to state a relation between the sides of DEF. For a right triangle with legs a and b the theorem states that its hypotenuse h is given by:
[tex]h^2=a^2+b^2[/tex]We can do the same for DEF. Its legs are DE and EF whereas its hypotenuse is DF so we get:
[tex]DF^2=DE^2+EF^2[/tex]And that's the equation requested in part B.
Find the area of this parallelogram. Be sure to include the correct unit in your answer.19 yd12 yd11 yd
Given a parallelogram as shown below:
The formula to calculate the area of the parallelogram is given to be:
[tex]A=b\times h[/tex]From the question provided, we have the following parameters:
[tex]\begin{gathered} a=12\text{ yd} \\ b=11\text{ yd} \\ h=9\text{ yd} \end{gathered}[/tex]Therefore, we can use the formula to calculate the area as shown below:
[tex]\begin{gathered} A=b\times h \\ A=11\times9 \\ A=99yd^2 \end{gathered}[/tex]The area of the parallelogram is 99 squared yards (99 yd²).