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2 years ago

Which sequence of transformations will verify that

Mathematics

1 answer:

AVprozaik [17]2 years ago

7 0

Answer:

C) △ABC is moved onto △A'B'C' after translating −16 units horizontally and 6 units vertically, and then reflecting across the x-axis.

Explanation: I did USA Test Prep.

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Rashid [163]

Answer:

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1 year ago

2. Triangle ABC and triangle EFD ARE ?PLEASE HELP

ad-work [718]

Answer:

similar SAS

Step-by-step explanation:

because the angle c has the angle c and sides are cb,ba .

because the angle d has the angle d and slide are df, Fe

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2 years ago

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Makovka662 [10]

Let $u=x^2-4$ and $v=4x-5$ . By the product rule,

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=\dfrac{\mathrm d(u^5)}{\mathrm dx}v^4+u^5\dfrac{\mathrm d(v^4)}{\mathrm dx}$

By the power rule, we have $(u^5)'=5u^4$ and $(v^4)'=4v^3$ , but $u,v$ are functions of $x$ , so we also need to apply the chain rule:

$\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}$

$\dfrac{\mathrm d(v^4)}{\mathrm dx}=4v^3\dfrac{\mathrm dv}{\mathrm dx}$

and we have

$\dfrac{\mathrm du}{\mathrm dx}=2x$

$\dfrac{\mathrm dv}{\mathrm dx}=4$

So we end up with

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3$

Replace $u,v$ to get everything in terms of $x$ :

$\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3$

We can simplify this by factoring:

$10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3=2(x^2-4)^4(4x-5)^3\bigg(5x(4x-5)+8(x^2-4)\bigg)$

$=2(x^2-4)^4(4x-5)^3(28x^2-57)$

7 0

2 years ago

Which set of ordered pairs that satisfy the equation

krok68 [10]

Answer:

Step-by-step explanation:

5 0

2 years ago

Solve 3x + 7 > -x - 9. Show each step of your work.

Kruka [31]

Answer:

x > -4

Step-by-step explanation:

Inequalities like these are solved the same way equations are solved, except when multiplying both sides you flip the less than/greater than sign:

3x + 7 > -x -9

3x + 7 - 7 > -x -9 -7 Subtract 7 from both sides

3x > -x -16

3x + x > -x -16 + x add x to both sides

4x > -16

$\frac{4x}{4}$ > $\frac{-16}{4}$ divide both sides by 4

x > -4

In this case the inequality sign did not have to be flipped since we never multiply both sides by a negative number.

3 0

2 years ago