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Allushta [10]
3 years ago
15

Dominic has $420 to spend at a bicycle store for some new gear and biking outfits. Assume all prices listed include taxes. He bu

ys a new bicycle vor $231.97. He buys 3 bicycle reflectors for $14.60 each anda pair ob bike gloves for$29.24. He plans to spend some or all of the money he has left to buy new biking outfirs for $48.25 each. What is the greatest number of outfits Dominic can buy with the money that's left over?
Mathematics
1 answer:
AnnyKZ [126]3 years ago
8 0

Answer: 2 is the greatest number of outfits he can buy.

Step-by-step explanation: no

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A cell phone plan costs $150 to start. Then there is a $25 charge each month.
grin007 [14]

Answer:

$175

Step-by-step explanation:

Since it’s only one month you just add 150 by 25

Correct me if I am wrong

7 0
3 years ago
NO LINKS PLZ AND REAL ANSWERS ASAP
ozzi
A & D are the correct answers!
8 0
3 years ago
Please help with 32 and 33, thank you.
PtichkaEL [24]

Answer:

32. ∛((x-7)/4) = f^(-1)(x)

33. -10x - 9

Step-by-step explanation:

32. We want to switch f(x) and x, and then solve for f(x) to get the inverse.

x = 4f(x)³ + 7

subtract 7 from both sides

x -7 = 4f(x)³

divide both sides by 4

(x-7)/4 = f(x)³

cube root both sides

∛((x-7)/4) = f(x)

make f(x) f^(-1)(x) because this is now the inverse

∛((x-7)/4) = f^(-1)(x)

the second answer is correct

33. for composition, we can treat (f · g) (x) as attached picture (content filter!), so we plug g(x) into f(x). This results in

2(-5x-7) + 5

expand

-10x - 14 + 5

add

-10x - 9

the second answer is correct

4 0
3 years ago
9.)Find the value of x. <br> A.) 75<br> B.) 70 <br> C.) 60 <br> D.) 80
Nikitich [7]
I think it's A but I'm not 100% sure
4 0
3 years ago
Read 2 more answers
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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