$\text{Given that }\$7000 \text{ is compounded semiannualy at a rate of }11\% \text{ for 21 years}\\
\\
\text{we know that the amount after t year when compounded is given by}\\
\\
A=P\left ( 1+\frac{r}{n} \right )^{nt}\\
\\
\text{here P is the principal amoount, so }P=7000,\\
\\
\text{r is the interest rate, }r=11\%=0.11\\
\\
\text{n is the number of times in a year, here semiannulay, so }n=2,\\
\\
\text{and t is the time, so }t=21\\
\\
\text{so the amount after 21 years is}$

$A=7000\left ( 1+\frac{0.11}{2} \right )^{2(21)}\\
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\Rightarrow A=7000\left ( 1+0.055 \right )^{42}\\
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\Rightarrow A=7000\left ( 1.055 \right )^{42}\\
\\
\Rightarrow A\approx 66328.68$

So the amount after 21 years is: $66328.68

7 0

3 years ago

In the diagram shown, m 1= 50 and m2 = (2x-20)

stepladder [879]

Answer:

x = 35

Step-by-step explanation:

Since m1=m2 :

$2x - 20 = 50 \\ 2x = 70 \\ x = 35$

6 0

2 years ago

. Use the vectors v = -i +2j and w= -3i + 4j to find 5v + 8w.

Anarel [89]

5v+8w

put values

5(-i+2j)+8(-3i+4j)

Do like you do algebraic addition

-5i+10j-24i+32j
-5i-24i+10j+32j
-29i+42j

Option A

5 0

2 years ago