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3 years ago

10

There is an ice cream and candy shop next door that is the same size as your bakery. How many thirtieths of the space does it ta

ke up? (Think Equivalent Fractions)

1 answer:

SSSSS [86.1K]3 years ago

8 0

Answer:

135,000 times

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one. Please have a look at the attached photo.

<u>We need to know the size of your bakery store. </u>

As we can see, the entire shopping center is: 45000 square feet and your shop is 1/10 of that space

=> Your shop area is: 45000*1/10 = 4500 square feet

In the question, they want to know how many thirtieths of the space does it take up, so let say thirtieths in numerical is: 1/30

Hence, the number of thirtieths it takes:

$\frac{Area}{Thirtieths}$ = $\frac{4500}{\frac{1}{30} }$ = 135,000 times

Hope it will find you well.

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A baker buys 19 apples of two different varieties to make pies. The total cost of the apples is $5.10. Granny Smith apples cost

Natasha2012 [34]

The baker bought 7 gala apples, and 12 granny smiths apples

<em><u>Solution:</u></em>

Let "x" be the number of gala apples bought

Let "y" be the number of granny smith apples bought

Cost of 1 gala apple = $ 0.30

Cost of 1 granny smith apple = $ 0.25

<em><u>A baker buys 19 apples of two different varieties to make pies</u></em>

Therefore,

number of gala apples bought + number of granny smith apples bought = 19

x + y = 19 --------- eqn 1

<em><u>The total cost of the apples is $5.10</u></em>

Therefore, we can frame a equation as:

number of gala apples bought x Cost of 1 gala apple + number of granny smith apples bought x Cost of 1 granny smith apple = 5.10

$x \times 0.30 + y \times 0.25 = 5.10$

0.3x + 0.25y = 5.1 -------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

From eqn 1,

x = 19 - y ---------- eqn 3

<em><u>Substitute eqn 3 in eqn 2</u></em>

0.3(19 - y) + 0.25y = 5.1

5.7 - 0.3y + 0.25y = 5.1

5.7 - 0.05y = 5.1

0.05y = 5.7 - 5.1

0.05y = 0.6

y = 12

<em><u>Substitute y = 12 in eqn 3</u></em>

x = 19 - 12

x = 7

Thus the baker bought 7 gala apples, and 12 granny smiths apples

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3 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

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$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
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$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

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Answer:

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