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ruslelena [56]
3 years ago
10

What is the answer to this question?

Mathematics
1 answer:
harina [27]3 years ago
4 0

Answer:

f(g(0) )= -1

Step-by-step explanation:

Find g(0) = -1

Then find f(-1) = 1

f(g(0) )= -1

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What is the interest rate of each investment if The annual interest on a $17,000 investment exceeds the interest earned on an $8
Butoxors [25]

<u>Solution-</u>

Let's assume, the rate of interest of $8000 is x%,

then the rate of interest of $17000 is (x+0.3x) =1.3x%

Interest earned by $8000,

i_1=\frac{8000\times x\times 1}{100} =80x

Interest earned by $17,000,

i_2=\frac{17000\times 1.3x\times 1}{100} =221x

According to the question,

\Rightarrow i_1=i_2+276

\Rightarrow 221x=80x+276

\Rightarrow 221x-80x=276

\Rightarrow 141x=276

\Rightarrow x=1.96

∴ Rate of interest of $8000 is 1.96% and rate of interest of $17000 is (1.3×1.96) =2.55%

3 0
3 years ago
57.4 divided by 10???????
Gnesinka [82]
57.4/10=5.74
the answer is 5.74

6 0
3 years ago
Read 2 more answers
Find the value of x. 55
Katena32 [7]

Answer:

90 - 55 = 35

its 35

Step-by-step explanation:

4 0
3 years ago
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Tim bought some 25 cent and some 29 cent stamps. He paid 7.60 dollars for 28 stamps. How many of each type of stamp did he buy?
Citrus2011 [14]

Answer:

25 cent stamps = 13 and 29 cent stamps = 15

Step-by-step explanation:

x = 25 cent stamps and y = 29 cent stamps

x + y = 28......x = 28 - y

0.25x + 0.29y = 7.60

0.25(28 - y) + 0.29y = 7.60

7 - 0.25y + 0.29y = 7.60

-0.25y + 0.29y = 7.60 - 7

0.04y = 0.60

y = 0.60 / 0.04

y = 15 <===== 29 cent stamps

x + y = 28

x + 15 = 28

x = 28 - 15

x = 13 <===== 25 cent stamps

lets check it...

0.25x + 0.29y = 7.60

0.25(13) + 0.29(15) = 7.60

3.25 + 4.35 = 7.60

7.60 = 7.60 (correct..it checks out)

8 0
3 years ago
Jayana is trying to run a certain number of miles by the end of the month. If Jayana is 30% of the way to achieving her goal and
torisob [31]
Jayana is trying to run 30 miles by the end of the months.

30% / 100% = 9 / x

9 * 100 = 900

900 / 30 = 30 miles
4 0
3 years ago
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