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tensa zangetsu [6.8K]

2 years ago

5

Find the perimeter????????

1 answer:

djverab [1.8K]2 years ago

4 0

Perimeter of triangle is calculated by adding up the three sides of the triangle.

In the given triangle ABC, vertices of triangle are given as $A(-2,2) B(0,4) C(1,-2)$

We calculate the length of side AB by the distance formula which states as :

For the points $(x_{1},y_{1}) and (x_{2},y_{2})$

distance = $\sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$

AB = $\sqrt{(0+2)^{2} + (4-2)^{2}} = 2\sqrt{2}$

BC= $\sqrt{(1-0)^{2} + (-2-4)^{2}} = \sqrt{37}$

AC= $\sqrt{(1+2)^{2} + (-2-2)^{2}} = 5$

Perimeter = AB+BC+AC = $=2\sqrt{2} + 5+ \sqrt{37}= 13.911$

So, Perimeter = 14 (Rounded to the nearest tenth)

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In Exercises 45–48, let f(x) = (x - 2)2 + 1. Match the<br> function with its graph

MA_775_DIABLO [31]

Answer:

45) The function corresponds to graph A

46) The function corresponds to graph C

47) The function corresponds to graph B

48) The function corresponds to graph D

Step-by-step explanation:

We know that the function f(x) is:

$f(x)=(x-2)^{2}+1$

45)

The function g(x) is given by:

$g(x)=f(x-1)$

using f(x) we can find f(x-1)

$g(x)=((x-1)-2)^{2}+1=(x-3)^{2}+1$

If we take the derivative and equal to zero we will find the minimum value of the parabolla (x,y) and then find the correct graph.

$g(x)'=2(x-3)$

$2(x-3)=0$

$x=3$

Puting it on g(x) we will get y value.

$y=g(3)=(3-3)^{2}+1$

$y=g(3)=1$

<u>Then, the minimum point of this function is (3,1) and it corresponds to (A)</u>

46)

Let's use the same method here.

$g(x)=f(x+2)$

$g(x)=((x+2)-2)^{2}+1$

$g(x)=(x)^{2}+1$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2x$

$0=2x$

$x=0$

Evaluatinf g(x) at this value of x we have:

$g(0)=(x)^{2}+1$

$g(0)=1$

<u>Then, the minimum point of this function is (0,1) and it corresponds to (C)</u>

47)

Let's use the same method here.

$g(x)=f(x)+2$

$g(x)=(x-2)^{2}+1+2$

$g(x)=(x-2)^{2}+3$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}+3$

$g(2)=3$

<u>Then, the minimum point of this function is (2,3) and it corresponds to (B)</u>

48)

Let's use the same method here.

$g(x)=f(x)-3$

$g(x)=(x-2)^{2}+1-3$

$g(x)=(x-2)^{2}-2$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}-2$

$g(2)=-2$

<u>Then, the minimum point of this function is (2,-2) and it corresponds to (D)</u>

<u />

I hope it helps you!

<u />

8 0

2 years ago

What is the mean and MAD of this data set?

just olya [345]

What data set???????

3 0

3 years ago

I need help on working this out. Can anyone help me out?

Tasya [4]

Gosh that looks hard. but i wish you the best of luck on it

8 0

2 years ago

Select the correct answer.

Sliva [168]

Answer:

C

Step-by-step explanation:

substitute X=7 in 2x+8, answer 22

substitute X=5, answer 18

option C

substituting X=7 in (X+4)+(X+4) also gives 22

substituting X=5 also gives 18

5 0

2 years ago

Seventy-two people are enrolled in the Bible learners department. If 37.5% are absent, how many people are absent?

Molodets [167]

192 people are absent

3 0

3 years ago