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3 years ago

Which answer is equal to ( sqrt 7)^3? A.7 3/2 B.7 2/3 C.7 1/6 D.7 1/3

2 answers:

OverLord2011 [107]3 years ago

7 0

$(\sqrt{7})^{3}$
$\sqrt{7}\sqrt{7}\sqrt{7}$
$\sqrt{7 * 7 * 7}$
$\sqrt{49 * 7}$
$\sqrt{343}$
$\sqrt{49 * 7}$
$\sqrt{49}\sqrt{7}$
$7\sqrt{7}$
$(7)^{\frac{3}{2}}$

The answer is A.

horrorfan [7]3 years ago

6 0

Try D As your Answer.

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A new car is purchased for $17,000 and over time its value depreciates by one half every 3 years. What is the value of the car 1

Jet001 [13]

Answer:

The answer is "$238".

Step-by-step explanation:

Current worth $= \$ 17,000$

depreciates by $\frac{1}{2}$ in 3 years.

time= 19 years

depreciates rate=?

Using formula:

$\to \text{Worth= Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}$

$\to A_t=A_0(1-\frac{r}{100})^t$

calculates depreciate value in 3 year $= \frac{1}{2} \times 17,000$

$= 8,500$

so,

$A_t=8,500\\\\A_0=17,000\\\\t=3\ years$

$\to A_t=A_0(1-\frac{r}{100})^t\\\\\to 8,500= 17,000(1-\frac{r}{100})^3\\\\\to \frac{8,500}{17,000}= (1-\frac{r}{100})^3\\\\\to \frac{1}{2}= (1-\frac{r}{100})^3\\\\\to (\frac{1}{2})^{\frac{1}{3}}= (1-\frac{r}{100})\\\\\to 0.793700526 = (1-\frac{r}{100})\\\\\to \frac{r}{100} = (1-0.793700526)\\\\\to \frac{r}{100} = (1-0.8)\\\\\to r= 0.2 \times 100 \\\\\to r= 20 \%$

depreciates rate= 20%

$\to \text{Worth= Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}$

$= \$ 17,000 (1- \frac{20}{100})^{19}\\\\= \$ 17,000 (1-0.2)^{19}\\\\= \$ 17,000 (0.8)^{19}\\\\= \$ 17,000 \times 0.014\\\\= \$ 238$

4 0

3 years ago

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What is x?<br><br> 3x - 4 + 1 = - 2x - 5 + 5x

ehidna [41]

3x - 4 + 1 = -2x - 5 + 5x

First, simplify 3x - 4 + 1 to 3x - 3. / Your problem should look like: 3x - 3 = -3x - 5 + 5x
Second, simplify -2x - 5 + 5x to 3x - 5. / Your problem should look like: 3x - 3 = 3x - 5
Third, cancel 3x on both sides. / Your problem should look like: -3 = -5
Fourth, since -3 = -5 is false, there is no solution.

Answer: No solution

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3 years ago

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What's the next number in the series. 65536, 256, 16..?

Artemon [7]