All categories

3 years ago

Given log3^2=0.631 and log3^7=1.771, what is log3^14?

Mathematics

1 answer:

Yuki888 [10]3 years ago

5 0

Answer:

3.442

Step-by-step explanation:

This means 10^(0.631) is 3^2, and 10^(1.771) is 3^7. Square the second equation on both sides to get 10^(1.771*2)=3^14, so the answer is 1.771*2 or 3.442.

You might be interested in

Brian’s kite is flying above a field at the end of 65m of string. If the angle of elevation to the kite measures 70; how high is

REY [17]

If the height is h, then h/65=sin70, so h=65sin70=61.08m approx.

7 0

3 years ago

What is the ratio of the intensities of an earthquake PP wave passing through the Earth and detected at two points 19 kmkm and 4

Tamiku [17]

Answer:

The ratio of the intensities is roughly 6:1.

Step-by-step explanation:

The intensity I() of an earthquake wave is given by:

$I = \frac{P}{4\pi d^{2}}$

<em>where P: is the power ans d: is the distance. </em>

Hence, the ratio of the intensities of an earthquake wave passing through the Earth and detected at two points 19 km and 46 km from the source is:

$\frac{I_{1}}{I_{2}} = \frac{P/4\pi d_{1}^{2}}{P/4\pi d_{2}^{2}}$

<em>where I₁ = P/4πd₁², d₁=19 km, I₁ = P/4πd₂² and d₂=46 km </em>

$\frac{I_{1}}{I_{2}} = \frac{d_{2}^{2}}{d_{1}^{2}}$

$\frac{I_{1}}{I_{2}} = \frac{46 km^{2}}{19 km^{2}}$

$\frac{I_{1}}{I_{2}} = 5.9:1$

Therefore, the ratio of the intensities is roughly 6:1.

I hope it helps you!

3 0

3 years ago

Lim x-1 x2 - 1/ sin(x-2)

balu736 [363]

Answer:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=0$

Explanation:

Assuming the correct expression is to find the following limit:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}$

Use the property the limit of the quotient is the quotient of the limits:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}$

Evaluate the numerator:

$\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}=\frac{1^2-1}{\lim_{x \to1}sin(x-2)}=\frac{0}{\lim_{x \to 1}sin(x-2}$

Evaluate the denominator:

Since $\lim_{x \to1}sin(x-2)\neq 0$

$\frac{0}{\lim_{x \to1}sin(x-2)}=0$

4 0

3 years ago

Plz answer these questions

Oxana [17]

Answer:

median 1B: 310

Step-by-step explanation:

4 0

3 years ago

The ratio of two complementary angles is 5 to 1. what are the angles?

densk [106]

Call x and y those angles. If they are complementary, then their sum equals 90°:

x + y = 90°

y = 90° - x

The ratio of them is 5 to 1. So,

x/y = 5/1

x/(90° - x) = 5/1

x = 5 * (90° - x)

x = 450° - 5x

x + 5x = 450°

6x = 450°

x = 450°/6

x = 75°

The measure of the other angle is

y = 90° - 75°

y = 15°

The angles are 75° and 15°.

I hope it helps. =)

6 0

3 years ago