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attashe74 [19]
3 years ago
15

Each pound of fruit costs $4. Write an expression that shows the total cost of the fruit. Use the variable you identified in que

stion 1. Btw the variable I used was "f".
Mathematics
1 answer:
vovangra [49]3 years ago
6 0

Answer:

$4f

Step-by-step explanation:

Let's assume f represents the number of pounds of fruit.

We need to multiply the number of pounds of fruit by the cost per pound.

This is $f * 4, or in a simpler form, $4f.

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neonofarm [45]

Answer:

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Step-by-step explanation

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7 0
3 years ago
V/-5+8=9 what does v equal
Sergeeva-Olga [200]
Subtract 8 from both sides and then multiply both sides by -5
V = -5
6 0
3 years ago
What is the solution for C when 10 C minus f equals -13 + c d
andreyandreev [35.5K]

Answer:

C would be equal to (f - 13)/(10 - d)

Step-by-step explanation:

In order to find this, you must manipulate the equation so that the left side has every term with a c in it. Then you can isolate it by dividing and find what it is equal to.

10c - f = -13 + cd -----> Add f to both sides

10c = f - 13 + cd -----> Subtract cd from both sides

10c - cd = f - 13 ------> Pull out c

c(10 - d) = f - 13 -----> Divide by (10 - d)

c = (f - 13)/(10 - d)

8 0
3 years ago
Harry is building a model car. The actual car has a length of 15 feet and tires with a radius of 24 inches. His model car has a
nasty-shy [4]

Answer:

B: two inches

Step-by-step explanation:

15/1.25=12

24/12=2

7 0
3 years ago
For the function​ below, find a formula for the upper sum obtained by dividing the interval [a comma b ][a,b] into n equal subin
Vlad [161]

Answer:

See below

Step-by-step explanation:

We start by dividing the interval [0,4] into n sub-intervals of length 4/n

[0,\displaystyle\frac{4}{n}],[\displaystyle\frac{4}{n},\displaystyle\frac{2*4}{n}],[\displaystyle\frac{2*4}{n},\displaystyle\frac{3*4}{n}],...,[\displaystyle\frac{(n-1)*4}{n},4]

Since f is increasing in the interval [0,4], the upper sum is obtained by evaluating f at the right end of each sub-interval multiplied by 4/n.

Geometrically, these are the areas of the rectangles whose height is f evaluated at the right end of the interval and base 4/n (see picture)

\displaystyle\frac{4}{n}f(\displaystyle\frac{1*4}{n})+\displaystyle\frac{4}{n}f(\displaystyle\frac{2*4}{n})+...+\displaystyle\frac{4}{n}f(\displaystyle\frac{n*4}{n})=\\\\=\displaystyle\frac{4}{n}((\displaystyle\frac{1*4}{n})^2+3+(\displaystyle\frac{2*4}{n})^2+3+...+(\displaystyle\frac{n*4}{n})^2+3)=\\\\\displaystyle\frac{4}{n}((1^2+2^2+...+n^2)\displaystyle\frac{4^2}{n^2}+3n)=\\\\\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12

but  

1^2+2^2+...+n^2=\displaystyle\frac{n(n+1)(2n+1)}{6}

so the upper sum equals

\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12=\displaystyle\frac{4^3}{n^3}\displaystyle\frac{n(n+1)(2n+1)}{6}+12=\\\\\displaystyle\frac{4^3}{6}(2+\displaystyle\frac{3}{n}+\displaystyle\frac{1}{n^2})+12

When n\rightarrow \infty both \displaystyle\frac{3}{n} and \displaystyle\frac{1}{n^2} tend to zero and the upper sum tends to

\displaystyle\frac{4^3}{3}+12=\displaystyle\frac{100}{3}

8 0
3 years ago
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