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Lera25 [3.4K]
3 years ago
11

How many time larger is 9x10^-8 than 3x10^-12

Mathematics
2 answers:
yaroslaw [1]3 years ago
4 0

Answer:

30000(thirty thousand) times larger

Step-by-step explanation:

9×10⁸ is the same as 900000000

3×10-¹² is the same as 0.000000000003

(9×10⁸)÷(3×10-¹²)

9÷3=3

10⁸÷10-¹²=10⁴

3×10⁴ which is 30000

kkurt [141]3 years ago
3 0

Step-by-step explanation:

\frac{9 \times 10 {}^{ - 8} }{3 \times 10 {}^{ - 12} }

Apply Quoteint of Powers. Divide the intergers. serpately.

Note that Quotient of Powers is

\frac{a {}^{x} }{a {}^{y} }  = a {}^{x  - y}

So the answer is

3 \times 10 {}^{ - 8 + 12}  = 3 \times 10 {}^{4}

The answer is

3 \times 10 {}^{4}

or

30000

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Which is greater 2 2/3, 2.45, 2 2/5
timofeeve [1]

This is least to greatest

2 2/5 or 2.40

2.45

2 2/3

2 2/5 , 2.45 , 2 2/3

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Solve the inequality –1 > –2(x – 4) – 5(4x – 7).
Ksivusya [100]

Answer:

x > 2

Step-by-step explanation:

–1 > –2(x – 4) – 5(4x – 7)

–1 > –2x  + 8 –20x + 35

–1 >  - 22x + 43

- 1 - 43 >  - 22x

- 44 >  - 22x

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3 years ago
Write three ratios equal to 5/40
AnnZ [28]
The answer is

<span>A. 2/14, 3/21, 4/28

B. 1/8, 2/16, 3/24

C. 1/9, 2/18, 3/27

D. 5/40, 5/45, 5/50</span>

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3 years ago
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Who Logarithmic equation is equivalent to the exponential equation below? e^4c=5
lesya692 [45]

Step-by-step explanation:

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3 years ago
99 POINT QUESTION, PLUS BRAINLIEST!!!
VladimirAG [237]
First, we have to convert our function (of x) into a function of y (we revolve the curve around the y-axis). So:


y=100-x^2\\\\x^2=100-y\qquad\bold{(1)}\\\\\boxed{x=\sqrt{100-y}}\qquad\bold{(2)} \\\\\\0\leq x\leq10\\\\y=100-0^2=100\qquad\wedge\qquad y=100-10^2=100-100=0\\\\\boxed{0\leq y\leq100}

And the derivative of x:

x'=\left(\sqrt{100-y}\right)'=\Big((100-y)^\frac{1}{2}\Big)'=\dfrac{1}{2}(100-y)^{-\frac{1}{2}}\cdot(100-y)'=\\\\\\=\dfrac{1}{2\sqrt{100-y}}\cdot(-1)=\boxed{-\dfrac{1}{2\sqrt{100-y}}}\qquad\bold{(3)}

Now, we can calculate the area of the surface:

A=2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\left(-\dfrac{1}{2\sqrt{100-y}}\right)^2}\,\,dy=\\\\\\= 2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=(\star)

We could calculate this integral (not very hard, but long), or use (1), (2) and (3) to get:

(\star)=2\pi\int\limits_0^{100}1\cdot\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\left|\begin{array}{c}1=\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\end{array}\right|= \\\\\\= 2\pi\int\limits_0^{100}\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\\\\\\ 2\pi\int\limits_0^{100}-2\sqrt{100-y}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\dfrac{dy}{-2\sqrt{100-y}}=\\\\\\

=2\pi\int\limits_0^{100}-2\big(100-y\big)\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\left(-\dfrac{1}{2\sqrt{100-y}}\, dy\right)\stackrel{\bold{(1)}\bold{(2)}\bold{(3)}}{=}\\\\\\= \left|\begin{array}{c}x=\sqrt{100-y}\\\\x^2=100-y\\\\dx=-\dfrac{1}{2\sqrt{100-y}}\, \,dy\\\\a=0\implies a'=\sqrt{100-0}=10\\\\b=100\implies b'=\sqrt{100-100}=0\end{array}\right|=\\\\\\= 2\pi\int\limits_{10}^0-2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=(\text{swap limits})=\\\\\\

=2\pi\int\limits_0^{10}2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4}\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^4}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^2}{4}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{\dfrac{x^2}{4}\left(4x^2+1\right)}\,\,dx= 4\pi\int\limits_0^{10}\dfrac{x}{2}\sqrt{4x^2+1}\,\,dx=\\\\\\=\boxed{2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx}

Calculate indefinite integral:

\int x\sqrt{4x^2+1}\,dx=\int\sqrt{4x^2+1}\cdot x\,dx=\left|\begin{array}{c}t=4x^2+1\\\\dt=8x\,dx\\\\\dfrac{dt}{8}=x\,dx\end{array}\right|=\int\sqrt{t}\cdot\dfrac{dt}{8}=\\\\\\=\dfrac{1}{8}\int t^\frac{1}{2}\,dt=\dfrac{1}{8}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\dfrac{1}{8}\cdot\dfrac{t^\frac{3}{2}}{\frac{3}{2}}=\dfrac{2}{8\cdot3}\cdot t^\frac{3}{2}=\boxed{\dfrac{1}{12}\left(4x^2+1\right)^\frac{3}{2}}

And the area:

A=2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx=2\pi\cdot\dfrac{1}{12}\bigg[\left(4x^2+1\right)^\frac{3}{2}\bigg]_0^{10}=\\\\\\= \dfrac{\pi}{6}\left[\big(4\cdot10^2+1\big)^\frac{3}{2}-\big(4\cdot0^2+1\big)^\frac{3}{2}\right]=\dfrac{\pi}{6}\Big(\big401^\frac{3}{2}-1^\frac{3}{2}\Big)=\boxed{\dfrac{401^\frac{3}{2}-1}{6}\pi}

Answer D.
6 0
4 years ago
Read 2 more answers
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