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3 years ago

(10^7)^2 as a single exponent

Mathematics

1 answer:

ycow [4]3 years ago

4 0

(10^7)^2
The answer = 10^14

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Find the angle of elevation of the sun from the ground to the top of a tree when a tree that is 10 yards tall casts a shadow 14

Hitman42 [59]

Answer:

The angle of elevation of the sun from the ground to the top of a tree is $36^{\circ}$ .

Step-by-step explanation:

As given

The sun from the ground to the top of a tree when a tree that is 10 yards tall casts a shadow 14 yards long.

Now by using the trignometric identity .

$tan\theta =\frac{Perpendicular}{Base}$

As figure is given below .

AB = Perpendicular = 10 yards

BC = Base = 14 yards

Putting all the values in the trignometric identity .

$tan\theta =\frac{AB}{BC}$

$tan\theta =\frac{10}{14}$

$\theta =tan^{-1}(\frac{10}{14})$

$\theta =36^{\circ}$

Therefore the angle of elevation of the sun from the ground to the top of a tree is $36^{\circ}$ .

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3 years ago

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I really need help please and thank you

loris [4]

Answer:

15 and 20

Step-by-step explanation:

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3 years ago

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Solve for x<br><br> C = 1/4πs^2x

kvasek [131]

Answer:

x=4C/πs^2

Step-by-step explanation:

C=1/4πs^2x

1) Divide both sides by π and s^2:

C/πs^2=1/4x

2) Multiply both sides by the reciprocal of 1/4: (4)

4C/πs^2=x

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2 years ago

If a =2, find 4a + 18

FinnZ [79.3K]

4×2+18

= 8+18

=26....

answer is 26..

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2 years ago

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A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago