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4 years ago

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Eddie sells homemade pies for $10 each. marsha sells pies for $8, and on one particular day, sells 3 more pies than eddie sells.

if they made the same amount of money today, how many pies did each of them sell?

1 answer:

serg [7]4 years ago

5 0

Eddie sold 12 pies and Marsha sold 15 pies.
They both made $120.
Find multiples of 8 that are divisible by 10

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Step-by-step explanation:

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3 years ago

The table below represents an exponential function what is the common ratio

Greeley [361]

Solution:

Given:

A table representing an exponential function.

The x-values represent the terms, while the y-values represent the numbers.

The common ratio is gotten from the numbers (y-values).

The formula for common ratio is given by;

$\begin{gathered} r=\frac{present\text{ term}}{preceed\text{ ing term}} \\ \\ \text{Hence,} \\ r=\frac{108}{36}=\frac{36}{12}=\frac{12}{4}=\frac{4}{1.33\ldots}=\frac{1.33\ldots.}{0.44\ldots} \\ r=3 \end{gathered}$

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2 years ago

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 $\dfrac{21}{16}$

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ss7ja [257]

Answer:

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If you need therapy or you're just bored, feel free to blow uf my notifications and spam in the comments. Or you can comment bcuz I got it wrong or something.

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3 years ago