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lianna [129]
3 years ago
9

Solve for x. Round your answer to the nearest tenth. (one decimal place) X=?? degrees

Mathematics
1 answer:
fredd [130]3 years ago
4 0

Answer:

<em>Angle x is congruent with the interior angle opposite side 8 (alternate interior angles)</em>

<u>Use tangent:</u>

  • tan x = 8/15
  • x = arctan (8/15)
  • x = 28.1° (rounded)
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I need help with this question
Bezzdna [24]

Answer:

It is all determined by how many spaces and which way the decimal point moves.

Step-by-step explanation:

So here are two examples

<em>Example 1:</em>

500000.00

To write this in scientific notation you would move the decimal place to the left.

You would move it 5 digits to the left so the exponent of 10 would be 5

5 * 10^{5}

<em>Example 2</em>

0.0000042

Here to write this in scientific notation you move the decimal place to the right.

You would move it 6 digits to the right so the exponent would be -6

4.2 * 10^{-6}

3 0
3 years ago
The pendulum of an antique clock has been damaged and needs to be replaced.If the original pendulum completed one swing every 1.
baherus [9]

Answer:

The length of pendulum is 55.853cm

Step-by-step explanation:

We are given

the original pendulum completed one swing every 1.5 seconds

so, T=1.5 sec

now, we can use time period formula

T=2\pi \sqrt{\frac{l}{g} }

l is the length of pendulum

g is acceleration due to gravity

g=9.8m/s^2

now, we can plug values

and then we can solve for l

1.5=2\pi \sqrt{\frac{l}{9.8} }

\frac{l}{9.8}=0.05699

l=0.55853m

In cm

l=55.853cm

3 0
3 years ago
Hiw do i get the exponents positive
Leokris [45]
Put them all under one. so it would b 1/everything negative
3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Csqrt%7B%28-3-4%29%5E%7B2%7D%2B%28-7-3%29%5E%7B2%29%20%7D" id="TexFormula1" title="\sqrt{(-3
bulgar [2K]

Answer:

Step-by-step explanation:

\sqrt{(-3-4)^2+(-7-3)^2

\sqrt{(-7)^2+(-10)^2

\sqrt{49+100

\sqrt{149

7 0
3 years ago
Let vector F = (6 x^2 y + 2 y^3 + 4 e^x) i + (7 e^{y^2} + 54 x) j . Consider the line integral of vector F around the circle of
balu736 [363]

Denote the circle of radius a by C. C is simple and closed, so by Green's theorem the line integral reduces to a double integral over the interior of C (call it D):

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_C(6x^2y+2y^3+4e^x)\,\mathrm dx+(7e^{y^2}+54x)\,\mathrm dy

=\displaystyle\iint_D\left(\frac{\partial(7e^{y^2}+54x)}{\partial x}-\frac{\partial(6x^2y+2y^3+4e^x)}{\partial y}\right)\,\mathrm dx\,\mathrm dy

=\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy

D is a circle of radius a, so we can write the double integral in polar coordinates as

\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^a(54-6r^2)r\,\mathrm dr\,\mathrm d\theta

a. For a=1, we have

\displaystyle\int_0^{2\pi}\int_0^1(54-6r^2)r\,\mathrm dr\,\mathrm d\theta=2\pi\int_0^1(54r-6r^3)\,\mathrm dr=\boxed{51\pi}

b. Let I(a) denote the integral with unknown parameter a,

I(a)=12\pi\int_0^a(9r-r^3)\,\mathrm dr\,\mathrm d\theta

By the fundamental theorem of calculus,

I'(a)=12\pi(9a-a^3)

I(a) has critical points when

12\pi(9a-a^3)=12\pi a(9-a^2)=0\implies a=0,a=\pm3

If a=0, then line integral is 0, so we ignore that critical point. For the other two, we would find I(\pm3)=243\pi.

8 0
3 years ago
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