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Drupady [299]
2 years ago
14

Consider the equation.

Mathematics
2 answers:
Lynna [10]2 years ago
8 0

1: -5/7

2: 26/7

With the help of mechzewd26

babunello [35]2 years ago
6 0

Answer:

If you use the addition property of equality, what number would you add to both sides of the equal sign to isolate the variable?

You would subtract 5/7 from both sides of the equation

What is the solution?

x = 26/7 or 3 and 5/7

Step-by-step explanation:

x + 5/7 = 31/7

Subtract 5/7 from both sides of the equation:

x + 5/7 - 5/7 = 31/7 - 5/7

x = 26/7

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Help me please! I'm so terrible at math
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3 years ago
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Solve: log(2)(3x + 8) = 5 Which is an equivalent equation?
Rzqust [24]

You can just simplify to make an equivalent equation:

(2)(3x + 8) = 5 ⇒ 6x + 16 = 5


Answer:

\boxed{\bf~6x + 16 = 5}


Hope it helped,


BioTeacher101


<em>(If you have any questions feel free to ask them in the comments)</em>


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3 years ago
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Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).
lutik1710 [3]

Answer:

The arc length is \dfrac{21}{16}

Step-by-step explanation:

Given that,

The given curve between the specified points is

x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}

The points from (\dfrac{9}{16},1) to (\dfrac{9}{8},2)

We need to calculate the value of \dfrac{dx}{dy}

Using given equation

x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}

On differentiating w.r.to y

\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})

\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})

\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})

\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}

We need to calculate the arc length

Using formula of arc length

L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}

Put the value into the formula

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}

L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}

L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}

L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}

L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}

L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}

Put the limits

L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})

L=\dfrac{21}{16}

Hence, The arc length is \dfrac{21}{16}

6 0
3 years ago
It was wartime when the Ricardos found out Mrs. Ricardo was pregnant. Ricky Ricardo was drafted and made out a? will, deciding t
ElenaW [278]

Answer:The girl will get $2,000, the mother $4,000 and the boy $8,000

Step-by-step explanation:

f the amount the girl gets = x

The mother gets twice as much as the girl = 2x

And the boy get twice as much as the mother = 4x


Overall

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7x = $14,000

Divide both sides by 7 so

x=$2,000

5 0
3 years ago
What is the formula for mortgages?
professor190 [17]

Answer:

Loan payment = Loan amount / Discount factor

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Step-by-step explanation:

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