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3 years ago

Can you guys please help me!!!!

Mathematics

1 answer:

iren2701 [21]3 years ago

3 0

Answer:

The graph of y = x^3 has been translated 3 units to the right and 4 units upward

Thanks and have a great day!

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Can some please help with this I will give brainliest.

3241004551 [841]

Answer:

$x^{-28}*7^{-56}$

Step-by-step explanation:

Firstly, you need to know that raising a fraction to a power raises both the top and the bottom to that power.

Second, a negative power on the top and bottom flips the fraction

Last, raising a power to a power multiplies the powers

So, $\frac{7^{-8^{7}}}{x^{4^7}}$ = $\frac{7^{-56}}{x^{28}}$ = $\frac{1}{x^{28}*7^{56}}$ = $x^{-28}*7^{-56}$

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2 years ago

"Which statement about angles is true? A. An angle is formed by two rays that do not have the same endpoint. B. An angle that tu

Amanda [17]

Answer:

Step-by-step explanation:

3 0

3 years ago

Solve the initial value problems:<br> 1/θ(dy/dθ) = ysinθ/(y^2 + 1); subject to y(pi) = 1

ladessa [460]

Answer:

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}$

Step-by-step explanation:

Given the initial value problem $\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\$ subject to y(π) = 1. To solve this we will use the variable separable method.

Step 1: Separate the variables;

$\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\\frac{1}{\theta}(\frac{dy}{sin\theta d\theta} ) =\frac{ y}{y^{2}+1 } \\\frac{1}{\theta}(\frac{1}{sin\theta d\theta} ) = \frac{ y}{dy(y^{2}+1 )} \\\\\theta sin\theta d\theta = \frac{ (y^{2}+1)dy}{y} \\integrating\ both \ sides\\\int\limits \theta sin\theta d\theta =\int\limits \frac{ (y^{2}+1)dy}{y} \\-\theta cos\theta - \int\limits (-cos\theta)d\theta = \int\limits ydy + \int\limits \frac{dy}{y}$

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y +C\\Given \ the\ condition\ y(\pi ) = 1\\-\pi cos\pi +sin\pi = \frac{1^{2} }{2} + ln 1 +C\\\\\pi + 0 = \frac{1}{2}+ C \\C = \pi - \frac{1}{2}$

The solution to the initial value problem will be;

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}$

5 0

3 years ago

How do you write 7.2% into decimal form?

LenaWriter [7]

You need to divide number by 100
7.2/100=0.072

5 0

3 years ago

Read 2 more answers

Simplify. Assume no variable is 0.

quester [9]

Step-by-step explanation:

Q16

$( {5x}^{3} {y}^{ - 5} )( {4xy}^{3} )$

First, we multiply the variable x together:

${x}^{3} \times x = {x}^{4}$

and remove brackets:

${5y}^{ - 5} \times {4x}^{4} \times {y}^{3}$

Now, we multiply the variable y:

${y}^{ - 5} \times {y}^{3} = {y}^{ - 2}$

Hence:

$5 \times 4 {x}^{4} \times {y}^{ - 2}$

Use the rule

${x}^{ - y} = \frac{1}{ {x}^{y} }$

Thus,

$5 \times 4 {x}^{4} \times \frac{1}{ {y}^{2} }$

Multiply:

$\frac{20 {x}^{4} }{ {y}^{2} }$

Q17

$( - 2 {b}^{3} c)(4 {b}^{2} {c}^{2} )$

$- 2 {b}^{3} c\times 4 {b}^{2} {c}^{2}$

Multiply the variable b together:

$- 2c \times 4 {b}^{5} {c}^{2}$

Multiply the variable c together:

$- 2 \times 4 {b}^{5} {c}^{3}$

Multiply.

$- 8 {b}^{5} {c}^{3}$

Q18:

$\frac{ {a}^{3} {n}^{7} }{a {n}^{4} }$

Divide the numerator and denominator by a:

$\frac{ {a}^{2} {n}^{7} }{ {n}^{4} }$

Apply the rule that

$\frac{ {x}^{y} }{ {x}^{z} } = {x}^{y - z}$

Hence, the equation will be

${a}^{2} {n}^{7 - 4}$

which is finally

${a}^{2} {n}^{3}$

Q19:

$\frac{ - {y}^{3} {z}^{5} }{ {y}^{2} {z}^{3} }$

Do the same rule said above:

$- \frac{ {z}^{5} {y}^{3 - 2} }{ {z}^{3} }$

$- \frac{y {z}^{5} }{ {z}^{3} }$

Do the rule again:

$- y {z}^{5 - 3}$

$- y {z}^{2}$

Q20:

$\frac{ - 7 {x}^{5} {y}^{5} {z}^{4} }{21 {x}^{7} {y}^{5} {z}^{2} }$

Cancel the common factor of 7:

$- \frac{ {x}^{5} {y}^{5} {z}^{4} }{3 {x}^{7} {y}^{5} {z}^{2} }$

$- \frac{ {y}^{5} {z}^{4} }{3 {y}^{5} {z}^{2} {x}^{7 - 5} }$

Thus,

$- \frac{ {y}^{5} {z}^{4} }{3 {y}^{5} {z}^{2} {x}^{2} }$

Cancel common factor of y^5

$- \frac{ {z}^{4} }{3 {z}^{2} {x}^{2} }$

$- \frac{ {z}^{4 - 2} }{3 {x}^{2} }$

$- \frac{ {z}^{2} }{3 {x}^{2} }$

Q21:

$\frac{9 {a}^{7} {b}^{5} {c}^{5} }{18 {a}^{5} {b}^{9} {c}^{3} }$

Cancel the common factor which is 9:

$\frac{{a}^{7} {b}^{5} {c}^{5} }{2 {a}^{5} {b}^{9} {c}^{3} }$

"Move" the variable a up using the rule:

$\frac{{a}^{7 - 5} {b}^{5} {c}^{5} }{2 {b}^{9} {c}^{3} }$

$\frac{{a}^{2} {b}^{5} {c}^{5} }{2 {b}^{9} {c}^{3} }$

"Move" the variable b down using the rule:

$\frac{{a}^{2} {c}^{5} }{2 {c}^{3} {b}^{9 - 5} }$

$\frac{{a}^{2} {c}^{5} }{2 {c}^{3} {b}^{4} }$

"Move" the variable c up using the rule

$\frac{{a}^{2} {c}^{5 - 3} }{2 {b}^{4} }$

$\frac{{a}^{2} {c}^{2} }{2 {b}^{4} }$

Q22:

$( {n}^{5} {)}^{4}$

Use the rule:

$({x}^{y} {)}^{z} = {x}^{yz}$

Hence,

${n}^{5 \times 4}$

${n}^{20}$

Q23:

$( {z}^{3} {)}^{6}$

Using the same rule above,

${z}^{3 \times 6}$

Therefore,

${z}^{18}$

5 0

3 years ago

Read 2 more answers