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slega [8]
3 years ago
13

The current value of an investment is 25% more than its initial value. The increase in value is $10 million. What was the initia

l value of the investment​
A.12.5 million
B.40 million
C.35 million
D.100 million
Mathematics
1 answer:
katrin [286]3 years ago
7 0
B is the answer ////////
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The correct answer above
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16e^2-88e+121<br> pls help me
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Step-by-step explanation:-8.8e +122

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A group of artists is selling small prints to raise money for charity. They sell p prints, and make a profit of f(p) on the sale
liraira [26]

The correct answer is is D) If the group sells 15 prints they will loose $85.

To figure out which statement is true, we have to evaluate the function ,

p(x)=17x-340 for all the values given in options (A)-(D).  A negative output represents a loss and  a positive output will represent a profit.

In A p=12 so f(12)=17\times 12-340=-136. In this case we gather that if they sell 12 prints, they will make a loss of $136. This tells us that option A is wrong.

In B, p=28 so f(28)=17\times 28-340=136. In this case we gather that if they sell 28 prints, they make a profit of $136. This tells us that option  B is wrong.

In C, p=35 so f(35)=17\times 35-340=225. In this case we gather that if they sell 35 prints, they make a profit of $225. This tells us that option  C is wrong.

Lastly, p=15 so f(15)=17\times 15-340=-85. In this case we gather that if they sell 28 prints, they make a loss of 85 dollars. From this we gather that D is the correct option.

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3 years ago
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ziro4ka [17]
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If this helped you please mark me brainliest :)
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6 0
3 years ago
Read 2 more answers
Given $m\geq 2$, denote by $b^{-1}$ the inverse of $b\pmod{m}$. That is, $b^{-1}$ is the residue for which $bb^{-1}\equiv 1\pmod
goblinko [34]

(2+3)^{-1}\equiv5^{-1}\pmod7 is the number <em>L</em> such that

5L\equiv1\pmod7

Consider the first 7 multiples of 5:

5, 10, 15, 20, 25, 30, 35

Taken mod 7, these are equivalent to

5, 3, 1, 6, 4, 2, 0

This tells us that 3 is the inverse of 5 mod 7, so <em>L</em> = 3.

Similarly, compute the inverses modulo 7 of 2 and 3:

2a\equiv1\pmod7\implies a\equiv4\pmod7

since 2*4 = 8, whose residue is 1 mod 7;

3b\equiv1\pmod7\implies b\equiv5\pmod7

which we got for free by finding the inverse of 5 earlier. So

2^{-1}+3^{-1}\equiv4+5\equiv9\equiv2\pmod7

and so <em>R</em> = 2.

Then <em>L</em> - <em>R</em> = 1.

6 0
3 years ago
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