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Oksi-84 [34.3K]

3 years ago

10

Find the area of the following geometric figure.

1 answer:

navik [9.2K]3 years ago

7 0

Find the area of a triangle with base of 6 m and altitude of 4 m.

12

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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

3 years ago

Which is bigger 0.6 or 2/3

lions [1.4K]

2/3 would be bigger
I hope that helps

8 0

3 years ago

Read 2 more answers

What is the result when 4x3 + 4x² + 13x + 6 is divided by 2x + 1?

Nitella [24]

Answer:

2x^2+x+6

Step-by-step explanation:

if the 4x3 meant 4x^3, then im pretty sure this is the correct answer

3 0

3 years ago

0.3y + y/z= y=10 and z=5

Xelga [282]

Answer:

32

Step-by-step explanation:

write the equation

0.3y + y/z=

Then you fill in what you know

0.3(10) + 10/5=

then we multiply/divide

30 + 10/5

then we continue to multiply/divide untill there is nothing to multiply/divide anymore

30+2

then we add/subtract

32

Our answer is 32

3 0

3 years ago

Line segment NY has endpoints N(-11, 5) and Y(3,-3).

777dan777 [17]

Given:

Line segment NY has endpoints N(-11, 5) and Y(3,-3).

To find:

The equation of the perpendicular bisector of NY.

Solution:

Midpoint point of NY is

$Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

$Midpoint=\left(\dfrac{-11+3}{2},\dfrac{5-3}{2}\right)$

$Midpoint=\left(\dfrac{-8}{2},\dfrac{2}{2}\right)$

$Midpoint=\left(-4,1\right)$

Slope of lines NY is

$m=\dfrac{y_2-y_1}{x_2-x_1}$

$m=\dfrac{-3-5}{3-(-11)}$

$m=\dfrac{-8}{14}$

$m=\dfrac{-4}{7}$

Product of slopes of two perpendicular lines is -1. So,

$m_1\times \dfrac{-4}{7}=-1$

$m_1=\dfrac{7}{4}$

The perpendicular bisector of NY passes through (-4,1) with slope $\dfrac{7}{4}$ . So, the equation of perpendicular bisector of NY is

$y-y_1=m_1(x-x_1)$

$y-1=\dfrac{7}{4}(x-(-4))$

$y-1=\dfrac{7}{4}(x+4)$

$y-1=\dfrac{7}{4}x+7$

Add 1 on both sides.

$y=\dfrac{7}{4}x+8$

Therefore, the equation of perpendicular bisector of NY is $y=\dfrac{7}{4}x+8$ .

6 0

3 years ago