6th grade math help me pleaseeee

Which statement about the graphs of h(x)=−2x+4 and k(x)=4x−4 is correct?

gayaneshka [121]

Answer:

A: The x-intercept of k(x) is half the x-intercept of h(x)

Step-by-step explanation:

Answer choices B, C, and D are concerned with slopes and y-intercepts. The coefficients of x in the functions are different and are related by a factor of -2, so the lines are not parallel, and one slope is not twice the other. The y-intercepts (constants) in each function are different, so they graphs do not cross the y-axis at the same place.

Hence answer choices B, C, and D can be eliminated.

The x-intercepts are found by setting y=0 and solving for x:

h(x) = 0 = -2x +4 ⇒ x = 4/2 = 2

k(x) = 0 = 4x -4 ⇒ x = 4/4 = 1

The x-intercept of k(x) is half that of h(x). . . . . . . matches choice A

7 0

3 years ago

Some one factor 4x^2+12x+8

Zinaida [17]

It stays the same right

5 0

3 years ago

Read 2 more answers

Please help me thank you

german

Step-by-step explanation:

y = 2x + 3

When x = 2

y = 2.2 + 3

y = 4 + 3

y = 7

When x = 0

y = 2.0 + 3

y = 0 + 3

y = 3

When x = -2

y = 2.(-2) + 3

y = -4 + 3

y = -1

When x = -4

y = 2.(-4) + 3

y = -8 + 3

y = -5

When x = -7

y = 2.(-7) + 3

y = -14 + 3

y = -9

8 0

3 years ago

R , S, and T are the vertices of one triangle. E, F, and D are the vertices of another triangle. mR = 60, mS = 80, mF = 60, mD =

Mumz [18]

Answer:

Yes, the two triangles are congruent, by ASA.

Step-by-step explanation:

In order to find if two triangles are congruent, you need the information of at least three out of the six incongnits of a triangle (the incognits are the three sides and the three angles)

Some of the ways to know if they're congruents are:

-ASA (Angle, side, angle): Two angles and the included side are equal for both triangles.

-AAS(Angle, angle, side): Two angles and a non-incuded side are equal for both triangles

-SAS(Side, angle, side): Two sides and the included angle are equal for both triangles.

In this case, two angles and one side of each triangle are known.

For the triangle RST, you know two angles and the included side.

For the triangle EFD, you know two angles and the non-included side. But is also known that the sum of the angles of a triangle is 180°, therefore you can obtain the third angle of EFD:

180=m∡F+m∡D+m∡E

m∡E= 180 - 60 -40

m∡E= 80°

Therefore:

m∡E=m∡S

m∡F=m∡R

RS=EF

Hence, two angles and the included side for RST and EFD are equal and the triangles are congruent by ASA.

8 0

3 years ago

Use matrices to determine the coordinates of the vertices of the reflected figure. Then graph the pre-image and the image on the

Troyanec [42]

Answer:

The coordinates of the vertices of the reflected figure are :

R' is (5 , -2) , S' is (3 , 5) , T' is (-7 , 6) ⇒ the right answer is (d)

Step-by-step explanation:

* When you reflect a point across the line y = x, the x-coordinate

and y-coordinate change their places.

- If the point is (x , y) then its image is (y , x)

* If you reflect over the line y = -x, the x-coordinate and y-coordinate

change their places and their signs

- If the point is (x , y) then its image is (-y , -x)

* Lets study the matrix of the reflection about the line y = x

- The matrix of the reflection about the line y = x is

$\left[\begin{array}{cc}0&1\\1&0\end{array}\right]$

- Because the x-coordinate and y-coordinate change places.

* Now lets solve the problem

- We will multiply the matrix of the reflection about y = x

by each point to find the image of each point

- The dimension of the matrix of the reflection about y = x

is 2×2 and the dimension of the matrix of each point is 2×1,

then the dimension of the matrix of each image is 2×1

∵ The point R is (-2 , 5)

∴ $R'=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]\left[\begin{array}{cc}-2\\5\end{array}\right]=$

$\left[\begin{array}{c}(0)(-2)+(1)(5)\\(1)(-2)+(0)(5)\end{array}\right]=\left[\begin{array}{c}5\\-2\end{array}\right]$

∴ R' is (5 , -2)

∵ The point S is (5 , 3)

∴ $S'=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]\left[\begin{array}{c}5\\3\end{array}\right]=$

$\left[\begin{array}{c}(0)(5)+(1)(3)\\(1)(5)+(0)(3)\end{array}\right]=\left[\begin{array}{c}3\\5\end{array}\right]$

∴ S' is (3 , 5)

∵ The point T is (6 , -7)

∴ $T'=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]\left[\begin{array}{c}6\\-7\end{array}\right]=$

$\left[\begin{array}{c}(0)(6)+(1)(-7)\\(1)(6)+(0)(-7)\end{array}\right]=\left[\begin{array}{c}-7\\6\end{array}\right]$

∴ T' is (-7 , 6)

* Lets look to the figures to find the right answer

∵ The R' is (5 , -2) , S' is (3 , 5) , T' is (-7 , 6)

∴ The right answer is (d)

4 0

4 years ago