You have 3/4 of a pound of peanuts. You have 10 cupcakes . What fraction of nuts is in each cupcake

1 answer:

8 0

Each cupcake would have 3/40 of a pound of nuts since that is 10x smaller than the total they have so they could put that amount into 10 cupcakes.

The answer is 3/40

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5 is what percent of 23

Eddi Din [679]

5 = what percent of 23

5 = x% of 23

5 = (x/100)*23

5 = 23x/100

5*100 = 23x

23x = 5*100

x = 5*100/23

x = 21.739

5 is ≈ 21.739% of 23

8 0

3 years ago

At George Washington High School the ratio of graduating seniors that go on to college, as compared to those who do not, is 7 :

wariber [46]

The solution to the problem is as follows:

<span>7(4)/4(4) = 28/16

28+16= 44
</span>
Therefore, the size of the graduating class would be 44 students in all.

I hope my answer has come to your help. God bless and have a nice day ahead!

7 0

3 years ago

A shopkeeper marks his goods at 40% above cost (the price he pays for the goods) and then offers a 25% discount. If a jacket cos

Y_Kistochka [10]

200(1.4)(.75)=$210.00

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2 years ago

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X^2-15x-108 Factoring need help

sveticcg [70]

x=15-657/2=15-373/2=-5.316

x=15+657/2=15+373=20.316

3 0

3 years ago

Find the equation of the line that is perpendicular to the line y = (-1/3)x -1 and passes through the point (1, 5)?

Anit [1.1K]

bearing in mind that perpendicular lines have negative reciprocal slopes, so

$\bf \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}~\hspace{10em}\stackrel{slope}{y=\stackrel{\downarrow }{-\cfrac{1}{3}}x-1} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{1}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{3}{1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{3}{1}\implies 3}}$

so we're really looking for a line whose slope is 3 and runs through (1,5)

$\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{5})~\hspace{10em} slope = m\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-5=3(x-1) \\\\\\ y-5=3x-3\implies y=3x+2$

4 0

2 years ago