Estimate the solution of a^2=21 to the nearest tenth

Please give the right answer and show work if possible

Step2247 [10]

Answer:

14

Step-by-step explanation:

Sin(∅) = Opposite angle ÷ hypotenuse.

Sin (45) = 7 $\sqrt{2}$ ÷ hypotenuse

hypotenuse = 7 $\sqrt{2}$ ÷ sin(45)

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Keep in mind

sin(45) = √2÷2

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so we have: Hypotenuse = 7√2 ÷ (√2÷2)

Hypotenuse= 14.

6 0

3 years ago

Cassandra and Yvette each have a calculator. Yvette starts at a 100 and subtracts 7 each time. Isandro starts at zero and adds 3

34kurt

This can be be solved the following way:
Yvette starts at 100 and subtracts 7 k times, as for Cassandra (or Isandro, the question is not clear on this) she starts at 0 and presses adds 3 k times. This leads to the following equation:
$100-7k=3k$
$100=10k
\\
10=k$
So for k=10 they will both display the same number.
We can now either choose the viewpoint of Cassandra or Yvette:
If we chose Yvette then we'll have $100-7*10=30$ and for Cassandra we have, $3*10=30$ .

8 0

3 years ago

Solve the following logarithmic equations. ln(32x^2) − 3 ln(2) = 3

Kazeer [188]

Answer:

$x=\pm \frac{\sqrt{e^{3} } }{2}$

Step-by-step explanation:

Rewrite the equation using the next propierties:

$log(\frac{1}{x} )=-log(x)$

$ylog(x)=log(x^{y} )$

$log(x*y)=log(x)+log(y)$

$ln(32x^{2} )-ln(2^{3})=3\\ ln(32x^{2} )+ln(\frac{1}{2^{3} } )=3\\ln(\frac{32x^{2} }{8})=3\\ ln(4x^{2} )=3$

Cancel logarithms by taking exp of both sides:

$e^{ln(4x^{2}) } =e^{3} \\4x^{2} =e^{3}$

Divide both sides by 4:

$x^{2} =\frac{e^{3} }{4}$

Take the square root of both sides:

$x=\pm \sqrt{\frac{e^{3} }{4} } =\pm \frac{\sqrt{e^{3} } }{2}$

6 0

3 years ago

1) I I ? 20- A) 999 C) 1219 B) 39° D) 1020

WITCHER [35]

Answer:

C

Step-by-step explanation:

$\angle{VUW} = 180 - 141 = 39$

$\angle{WVU} = 180 - (20 + 39) = 121$

Answer: 121

8 0

3 years ago

State the horizontal asymptote of the rational function. f(x) = quantity x squared plus three x minus two divided by quantity x

omeli [17]

$f(x)=\dfrac{x^2+3x-2}{x-2}\\ D_f=\mathbb{R}\setminus\{2\}\\\\ \displaystyle \lim_{x\to \infty}\dfrac{x^2+3x-2}{x-2}=\lim_{x\to \infty}\dfrac{x\left(x+3-\dfrac{2}{x}\right)}{x\left(1-\dfrac{2}{x}\right)}=\lim_{x\to \infty}\dfrac{x+3-\dfrac{2}{x}}{1-\dfrac{2}{x}}=\dfrac{\infty}{1}=\infty\\ \lim_{x\to -\infty}\dfrac{x^2+3x-2}{x-2}=\lim_{x\to -\infty}\dfrac{x\left(x+3-\dfrac{2}{x}\right)}{x\left(1-\dfrac{2}{x}\right)}=\lim_{x\to -\infty}\dfrac{x+3-\dfrac{2}{x}}{1-\dfrac{2}{x}}=\dfrac{-\infty}{1}=-\infty$

The limits are not numbers, so the asymptote doesn't exist.

8 0

3 years ago