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4 years ago

Triangle ABC and its transformation DEF are shown. What transformation of triangle ABC produced triangle DEF?

Mathematics

2 answers:

GenaCL600 [577]4 years ago

8 0

Answer:

Step-by-step explanation:

I Promise

SVETLANKA909090 [29]4 years ago

4 0

Answer: a

Step-by-step explanation:

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Can someone plz help me on these 2

kodGreya [7K]

Answer:

m=5.95

m=3

Step-by-step explanation:

4 0

3 years ago

Use part 1 of the fundamental theorem of calculus to find the derivative of the function. g(x) = x t3 t5 dt 0

salantis [7]

Using the fundamental theorem of calculas the derivative of function g(x)= $xt^{3}+t^{5}$ at x=0 is $t^{3}$ .

Given a function g(x)= $xt^{3}+t^{5}$ .

We are required to find the derivative of the function g(x) at x=0.

Function is relationship between two or more variables expressed in equal to form. The values entered in a function are part of domain and the values which we get from the function after entering of values are part of codomain of function. Differentiation is the sensitivity to change of the function value with respect to a change in its variables.

g(x)= $xt^{3}+t^{5}$

Differentiating with respect to x.

d g(x)/dx= $t^{3}$ +0 [Differentiation of x is 1 and differentiation of constant is 0]

= $t^{3}$

Hence using the fundamental theorem of calculas the derivative of function g(x)= $xt^{3}+t^{5}$ at x=0 is $t^{3}$ .

The function given in the question is incomplete. The right function will be g(x)= $xt^{3}+t^{5}$ .

Learn more about differentiation at brainly.com/question/954654

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5 0

2 years ago

Thank you so much to anyone who answers this question!

Anna71 [15]

C. The 40th customer

4 0

3 years ago

Drag the tiles to the correct boxes to complete the pairs.

Elza [17]

Answer:

Part 1) -7 → $\dfrac{y^2-y+6}{-2(y+7)}$

Part 2) 3/2 → $\dfrac{y^2-2y+1}{2y-3}$

Part 3) 1 → $\dfrac{5y^2-6y+1}{-5(y-1)}$

Part 4) -1/4 → $\dfrac{y(y+5)}{4y+1}$

Step-by-step explanation:

<u><em>The complete question in the attached figure</em></u>

we know that

To find out the non permissible replacements for y, equate the denominator of each expression equal to 0.

step 1

we have

$\dfrac{y^2-2y+1}{2y-3}$

Equate (2y-3) equal to 0.

$2y-3=0$

solve for y

Adds 3 both sides.

$2y=3$

Divide both sides by 2.

$y=\dfrac{3}{2}$

therefore

3/2 is the non permissible replacement for y.

step 2

we have

$\dfrac{y(y+5)}{4y+1}$

Equate (4y+1) equal to 0.

$4y+1=0$

Subtract 1 both sides

$4y=-1$

Divide by 4 both sides

$y=-\dfrac{1}{4}$

therefore

-1/4 is the non permissible replacement for y.

step 3

we have

$\dfrac{5y^2-6y+1}{-5(y-1)}$

Equate -5(y-1) equal to 0.

$-5(y-1)=0$

Divide by -5 both sides

$y-1=0$

Adds 1 both sides

$y=1$

therefore

1 is the non permissible replacement for y.

step 4

we have

$\dfrac{y^2-y+6}{-2(y+7)}$

Equate -2(y+7) equal to 0.

$-2(y+7)=0$

Divide by -2 both sides

$y+7=0$

Subtract 7 both sides

$y=-7$

therefore

-7 is the non permissible replacement for y

7 0

3 years ago

I need help with this

vazorg [7]

Zx = 5g(2x - c) Use the Distributive Property
zx = 10gx -5cg Subtract 10gx from both side
zx - 10gx = -5cg Factor out the x
x (z - 10g) = -5cg Divide both sides by (z-10g)
X= -5cg/z-10g

Hope this helps and please mark me as brainlest and like:)

6 0

3 years ago