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andreyandreev [35.5K]

3 years ago

6

The product of x and 7

2 answers:

solmaris [256]3 years ago

7 0

The product of x and 7.

7x

Answer: 7x

Natasha_Volkova [10]3 years ago

3 0

The answer that you are looking for is 7x
The product of x and 7 is 7x
have a good day!

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Compound interest of $1000 is invested at 10% compounded continuously, the future value s at any time t in years is given by s=1

crimeas [40]

Answer: 6.93 years

Step-by-step explanation:

Given

Rate of interest is $r=10\%$

Future value is given by $s=1000e^{0.1t}$

For the investment to double itself, i.e. $s=2000$

$\Rightarrow 2000=1000e^{0.1t}\\\Rightarrow 2=e^{0.1t}\\\\\text{Taking log both sides}\\\\\Rightarrow \ln 2=0.1t\\\\\Rightarrow t=\dfrac{\ln 2}{0.1}\\\\\Rightarrow t=6.93\ \text{years}$

It takes around 6.93 years to double the investment.

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3 years ago

Which fraction has a terminating decWhich is the decimal expansion of StartFraction 7 Over 22 EndFraction? ModifyingAbove 0.318

12345 [234]

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17?15

Step-by-step explanation:

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4 years ago

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Indicate the answer choice that best completes the statemer

Troyanec [42]

Answer:

25

Step-by-step explanation:

-25 to 0 is 25 and 25 to 0 is 25

Opposite of -25 is 25

So lettter d

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3 years ago

Translate each inequality. ¨a minimum of 10¨

Alexandra [31]

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X ≥ 10

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3 years ago

Let u = (1, k) and v = (2, 1). Find k such that The distance between u and v is 1 u and v are orthogonal The angle between u and

SVETLANKA909090 [29]

Answer:k=1,k=-2,k= $8\pm 5\sqrt{3}$

Step-by-step explanation:

Given two vectors

$u=1\hat{i}+k\hat{j}$

$v=2\hat{i}+1\hat{j}$

$\left ( i\right )$ Distance between them is given by

$|u-v|=\sqrt{\left ( 2-1\right )^2+\left ( 1-k\right )^2}=1$

squaring both side

$1^{2}+\left ( 1-k\right )^2=1$

$k^2-2k+1=0$

$\left ( k-1\right )^2=0$

k=1

$\left ( ii\right )$

angle between u and v is 90 i.e. orthogonal

$u\dot v=0$

$\left ( 1\hat{i}+k\hat{j}\right )\dot \left ( 2\hat{i}+1\hat{j}\right )$ =0

2+k=0

k=-2

$\left ( iii\right )$

angle between u & v is $\frac{\pi }{3}$

$u\dot v=|u||v|cos\left (\frac{\pi }{3}\right )$

$|u|=\sqrt{1^2+k^2}$

$|v|=\sqrt{2^2+1^2}$

$2+k=\left ( \sqrt{1+k^2}\right )\left ( \sqrt{5}\right )cos\left ( \frac{\pi }{3}\right )$

$\left ( 4+2\right )^2=\left ( 1+k^2\right )5$

$k^2-16k-11$ =0

k= $8\pm 5\sqrt{3}$

7 0

3 years ago