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1 year ago

Destiny just received two separate gifts from her great-great-grandmother.

Mathematics

2 answers:

hichkok12 [17]1 year ago

5 0

answer: 6 is the greatest number of snack bags that Destiny can make. Step-by-step

Step-by-step explanation:

The first gift is a box of 18 chocolate candy bars, and the second gift is a pack of 12 cookies.

lubasha [3.4K]1 year ago

5 0

Answer:

There are 6 snack bags that Destiny can make.

Step-by-step explanation:

Since we have given that

Number of chocolate candy bars in the first gift =18

Number of cookies in the second gift = 12

Since Destiny wants to use all of the chocolate candy bars and cookies to make identical snack bags for her cousins,

So, Destiny can make the greatest number of snack bags is given by

So, there are 6 snack bags that Destiny can make.

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Answer:

y = 3/8x

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Step-by-step explanation:

Since we are only making 1/2 of the full batch of muffins, we only need to use 1/2 the cups of sugar:

$\frac{3}{4} (\frac{1}{2} )= \frac{3}{8}$ cups of sugar.

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2 years ago

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Gabe and Eugenia bought a house! Their loan is for $117,000 , for 15 years at an annual interest rate of 4% . This results in a

cupoosta [38]

Answer:

Principal element is $475.43

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Step-by-step explanation:

The amount of interest paid in month one is 4%*$117,000*1/12=$390

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The principal element of monthly payment is the monthly payment minus interest.

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Ultimately,$475.43 goes toward reducing her loan balance while the $390 is interest on loan

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1 year ago

You have two circles, one with radius r and the other with radius R. You wish for the difference in the areas of these two circl

gtnhenbr [62]

Answer:

maximum difference of radii = $(r-R)=\frac{1}{2\pi ^{2}}$

Step-by-step explanation:

We know that area of circle is given by

$A=\pi \times (radius)^{2}$

For circle with radius 'r' we have

$A_{1}=\pi \times (r)^{2}$

For circle with radius 'R' we have

$A_{2}=\pi \times (R)^{2}$

Now according to given condition we have

$A_{1}-A_{2}\leq \frac{5}{\pi }$

$\Rightarrow \pi r^{2}-\pi R^{2}\leq \frac{5}{\pi }\\\\\Rightarrow (r^{2}-R^{2})\leq \frac{5}{\pi ^{2}}\\\\(r+R)(r-R)\leq \frac{5}{\pi ^{2}}\\\\\because (a^{2}-b^{2})=(a+b)(a-b)\\\\(r+R)=10(Given)\\\\\Rightarrow(r-R)\leq \frac{5}{10\pi ^{2}}\\\\\therefore (r-R)\leq\frac{1}{2\pi ^{2}}$