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3 years ago

In triangle ABC, ∠A=120°, a=3, and b=2. Find angle ∠C to the nearest degree.

Mathematics

1 answer:

sweet [91]3 years ago

6 0

Answer:

Step-by-step explanation:

Use Law of Sines to find ∠B.

∠B = arcsin(b·sinA/a) ≅ 35°

∠C = 180°-∠A-∠B = 25°

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2 3<br> —- = —-<br> x + 2 2x - 4

Ierofanga [76]

Answer:

23x-4

Step-by-step explanation:

x+22x-4 ( If a <u>term</u> doesn't have a coefficient, it is considered that the coefficient is 1

1x+22x-4 Collect <u>like terms</u> by adding their coefficients (1+22)x-4

Then, add the numbers.

(1+22)x-4

23x-4

8 0

3 years ago

find a curve that passes through the point (1,-2 ) and has an arc length on the interval 2 6 given by 1 144 x^-6

taurus [48]

Answer:

$f(x) = \frac{6}{x^2} -8$ or $f(x) = -\frac{6}{x^2} + 4$

Step-by-step explanation:

Given

$(x,y) = (1,-2)$ --- Point

$\int\limits^6_2 {(1 + 144x^{-6})} \, dx$

The arc length of a function on interval [a,b]: $\int\limits^b_a {(1 + f'(x^2))} \, dx$

By comparison:

$f'(x)^2 = 144x^{-6}$

$f'(x)^2 = \frac{144}{x^6}$

Take square root of both sides

$f'(x) =\± \sqrt{\frac{144}{x^6}}$

$f'(x) = \±\frac{12}{x^3}$

Split:

$f'(x) = \frac{12}{x^3}$ or $f'(x) = -\frac{12}{x^3}$

To solve fo f(x), we make use of:

$f(x) = \int {f'(x) } \, dx$

For: $f'(x) = \frac{12}{x^3}$

$f(x) = \int {\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = \frac{12}{2x^2} + c$

$f(x) = \frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = \frac{6}{1^2} + c$

$-2 = \frac{6}{1} + c$

$-2 = 6 + c$

Make c the subject

$c = -2-6$

$c = -8$

$f(x) = \frac{6}{x^2} + c$ becomes

$f(x) = \frac{6}{x^2} -8$

For: $f'(x) = -\frac{12}{x^3}$

$f(x) = \int {-\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = -\frac{12}{2x^2} + c$

$f(x) = -\frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = -\frac{6}{1^2} + c$

$-2 = -\frac{6}{1} + c$

$-2 = -6 + c$

Make c the subject

$c = -2+6$

$c = 4$

$f(x) = -\frac{6}{x^2} + c$ becomes

$f(x) = -\frac{6}{x^2} + 4$

3 0

3 years ago

HELP ASAP!! questions are in the pic

garik1379 [7]

Answer:

1: 13 and 12

2: 9 and 10

3: 8 and 9

4: 5 and 6

5: 7 and 8

6: 4 and 5

Step-by-step explanation:

Hope this helps.

6 0

3 years ago

Read 2 more answers

A model rocket is launched with an initial upward velocity of 57m/s. The rocket's height h (in meters) after t seconds is given

vesna_86 [32]

The values of for which the rocket's height is 29 meters are 0.75 and -0.67.

<h3>What is a velocity?</h3>

Velocity can be described as a vector quantity having both direction and magnitude which is the distance traveled per time.

We were given the rocket's height h as $h=57t-5t^2$ which is an equation, and we were been given the height which is the value that can be input into the equation as is 29 meters , then we can proceed to simplify the equation as

$h=57t-5t^2$

29=57t-5t^2

then we will have $57t-5t^2-29=0$

with the use of quadratic formula the values of t are : 0.75 and -0.67

Then we can chose 0.75 seconds

Therefore, after the calculation with the use of the quadratic equation, we can come into conclusion that the values that are required in the question are 0.75 and -0.67.

Learn more about velocity at:

brainly.com/question/4535155

#SPJ1

NOTE: This is a complete question, there were no given options, even on the internet.

6 0

1 year ago

A rectangular safe can hold 7,296 cubic inches. Gold bars that are 4 inches by 8 inches by 2 inches fit to completely fill

Eddi Din [679]

Answer:

your answer is 114

I hope it's helps you

7 0

3 years ago