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polet [3.4K]
4 years ago
8

I need help with 26 ASAP plzz

Mathematics
1 answer:
rewona [7]4 years ago
4 0
The answer to your question is A 35 degrees.
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HELP!
MAVERICK [17]

Answer: Hence, this investment would be worth of $366.756

Step-by-step explanation:

Since we have given that

Amount he invests = $350

Rate of interest compounded quarterly = 1.5%

Number of years = 50 years = 1.25 quarters

As we know the formula for "Compound Interest ":

Amount=P(1+\frac{r}{400})^n\\\\Amount=350(1+\frac{1.5}{400})^{12}\\\\Amount=\$366.07

For remaining half year, we first find the interest with using the above amount as principal amount.

Interest=\frac{366.07\tiems 1.5\times 0.5}{4\times 100}\\\\Interest=\$0.686

Hence, this investment would be worth of

366.07+0.686\\\\=\$366.756

4 0
3 years ago
Read 2 more answers
Simplify the expression. (81x^12/3x^3)^1/3
ser-zykov [4K]
Raise each number to the power outside the parentheses.

\left( \dfrac{81x^{12}}{3x^3} \right) ^{ \frac{1}{3} } = \\

= \left( 27x^{12-3} \right) ^{ \frac{1}{3} } = \\

= \left( 27x^{9} \right) ^{ \frac{1}{3} } = \\

= \left( 3^3x^{9} \right) ^{ \frac{1}{3} } = \\

= \left( 3^3 \right) ^{ \frac{1}{3} } \times \left( x^{9} \right) ^{ \frac{1}{3} } \\

= \left( 3^{3 \times \frac{1}{3} } \right)  \times \left( x^{9 \times  \frac{1}{3} }} \right)  \\

= 3^1 x^3 \\

= 3x^3



6 0
3 years ago
The temperature on a winter day increased 37oF. If the beginning temperature was -9oF, what was the temperature after the increa
Rashid [163]
If you would like to know what was the temperature after the increase, you can calculate this using the following step:

the beginning temperature: -9
increased: 37
the temperature after the increase: -9 + 37 = 28oF

The temperature after the increase was 28oF.
5 0
4 years ago
Read 2 more answers
2 3 + 6^{2} ÷ 9 i need help with this one
creativ13 [48]

Answer: 7

Step-by-step explanation:

3+6^(2) / 9

3 + 36 / 9

3   +   4

    7

5 0
3 years ago
Hello pleasee!!!!!!!!!!!!
Vikentia [17]
It’s is c because the the dom have a goodie package that you can do that for me
7 0
3 years ago
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