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2 years ago

Which rule describes the x-coordinates in the translation below?.

Mathematics

2 answers:

rosijanka [135]2 years ago

6 0

9514 1404 393

Answer:

(a) x + 0

Step-by-step explanation:

When translation is up or down, the x-coordinate doesn't change. It will be ...

x + 0

Stels [109]2 years ago

6 0

Answer:x+0

Step-by-step explanation:I HAVE DONE THIS BEFORE

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A pair of shoes is on sale for $66.00. This is 75% of the original price. What was the original price?

sergejj [24]

Answer:

$88.00

Step-by-step explanation:

66=3/4x

66/.75=66*4/3=88

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2 years ago

Please help will give brainliest

Ksju [112]

Answer:

a² = 144

b² = 25

c² = 169

c = 13

Step-by-step explanation:

• a² = 12×12

= 144

• b² = 5×5

= 25

• a² + b² = c²

= 144 + 25 = 169

• c = _/169

= 13

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2 years ago

Solve the equation. StartFraction dx Over dt EndFraction equals StartFraction 1 Over x e Superscript t plus 7 x EndFraction An i

MAVERICK [17]

Answer:

$\frac{1}{49}(7x-1)e^{7x}+e^{-t}=C$

Step-by-step explanation:

We are given that

$\frac{dx}{dt}=\frac{1}{xe^{t+7x}}$

We have to find the implicit function

Using separation variable method

$\frac{dx}{dt}=\frac{1}{xe^t\cdot e^{7x}}$

By using property $x^a\cdot x^y=x^{a+y}$

$xe^{7x}dx=e^{-t}dt$

By using property $\frac{1}{x^a}=x^{-a}$

Taking integration on both sides

$\int xe^{7x}dx=\int e^{-t}dt$

Parts integration method

$\int u\cdot v dx=u\int vdx-\int (\frac{du}{dx}\int vdx)dx$

By parts integration method

$x\int e^{7x}dx-\int (\frac{dx}{dx}\int e^{7x}dx)dx=-e^{-t}+C$

Using formula $\int e^{ax} dx=\frac{e^{ax}}{a}+C$

$\frac{xe^{7x}}{7}-\frac{1}{7}\int e^{7x}dx=-e^{-t}+C$

$\frac{xe^{7x}}{7}-\frac{1}{49}e^{7x}+e^{-t}=C$

$\frac{1}{49}(7x-1)e^{7x}+e^{-t}=C$

We are given that

$F(x,t)=C$

$F(x,t)=\frac{1}{49}(7x-1)e^{7x}+e^{-t}=C$

8 0

3 years ago

Simplify fraction 4/28 and 24/80

Degger [83]

Answer:

1. 1/7

2. 3/10

Step-by-step explanation:

4/28 can be divided by 4

24/80 can be divided by 8

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2 years ago

Giving brainliestttt

Nataly [62]

Answer:

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2 years ago