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3 years ago

BRAINLIEST!!!! HELP!!

Mathematics

1 answer:

dusya [7]3 years ago

4 0

Answer:

Step-by-step explanation:

Consider the right triangle with hypotenuse of 60 yards and the height 12 being the side opposite the angle of incline, say x°, then

sinx° = $\frac{opposite}{hypotenuse}$ = $\frac{12}{60}$ , thus

x = $sin^{-1}$ ( $\frac{12}{60}$ ) ≈ 11.54°

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How can you use similar triangles to find a missing measurement?

Vlad1618 [11]

Answer:

By comparison of the sides.

Step-by-step explanation:

If triangles are similar, that means their angles are equal and sides are proportional.

If you already have a proportion of two corresponding sides between triangles, then use that proportion and set it equal to a side you already know the measurement to. Hopefully that side is corresponding to your missing measurement, and you can then solve it.

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2 years ago

Evaluate cos 300' without using a calculator, 2 O A. 1 / O B. 1 о c va

storchak [24]

Answer:

$Cos \ 300 = \dfrac{1}{2}$

Step-by-step explanation:

$\sf \boxed{Cos \ (2\pi - \theta) = Cos \ \theta}$

Cos 300 = Cos ( 360 - 60)

= Cos 60

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2 years ago

Rachel is saving money to buy an outfit for a party she has already saved $17 she is earning $5 each month for being a mother's

AveGali [126]

Answer:

18 Weeks

Step-by-step explanation:

$39.50=17 + (5*x)

$39.50-17=17-17 + (5*x)

$22.5/5=(5*x)/5

It will take 4.5 months to save up. If each month has 4 weeks it will take 18 weeks

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2 years ago

A brine solution of salt flows at a constant rate of 4 L/min into a large tank that initially held 100 L of pure water. The solu

frutty [35]

Answer:

a. $m(t) = 26.67 - 26.67e^{-0.03t}$ b. 7.44 s

Step-by-step explanation:

a. If the concentration of salt in the brine entering the tank is 0.2 kg/L, determine the mass of salt in the tank after t min.

Let m(t) be the mass of salt in the tank at any time, t.

Now, since a brine solution flows in at a rate of 4 L/min and has a concentration of 0.2 kg/L, the mass flowing in per minute is m' = 4 L/min × 0.2 kg/L = 0.8 kg/min

Now, the concentration in the tank of volume 100 L at any time, t is m(t)/100 L. Since water flows out at a rate of 3 L/min, the mass flowing out per minute is

m(t)/100 × 3 L/min = 3m(t)/100 kg/min

Now the net rate of change of mass of salt in the tank per minute dm/dt = mass flowing in -mass flowing out

dm/dt = 0.8 kg/min - 3m(t)/100 kg/min

So, dm/dt = 0.8 - 0.03m(t)

The initial mass of salt entering m(0) = 0 kg

dm/dt = 0.8 - 0.03m(t)

separating the variables, we have

dm/[0.8 - 0.03m(t)] = dt

Integrating, we have

∫dm/[0.8 - 0.03m(t)] = ∫dt

-0.03/-0.03 × ∫dm/[0.8 - 0.03m(t)] = ∫dt

1/(-0.03)∫-0.03dm/[0.8 - 0.03m(t)] = ∫dt

-1/0.03㏑[0.8 - 0.03m(t)] = t + C

㏑[0.8 - 0.03m(t)] = -0.03t - 0.03C

㏑[0.8 - 0.03m(t)] = -0.03t + C' (C'= -0.03C)

taking exponents of both sides, we have

$0.8 - 0.03m(t) = e^{-0.03t + C'} \\0.8 - 0.03m(t) = e^{-0.03t}e^{C'}\\0.8 - 0.03m(t) = Ae^{-0.03t} A = e^{C'}\\0.03m(t) = 0.8 - Ae^{-0.03t}\\m(t) = 26.67 - \frac{A}{0.03} e^{-0.03t}\\when t = 0 \\m(0) = 0\\m(0) = 26.67 - \frac{A}{0.03} e^{-0.03(0)}\\\\0 = 26.67 - \frac{A}{0.03} e^{0}\\26.67 = \frac{A}{0.03} \\\frac{A}{0.03} = 26.67\\\frac{A}{0.03} = 6.67\\A = 26.67 X 0.03\\A = 0.8\\m(t) = 26.67 - \frac{A}{0.03} e^{-0.03t}\\\\m(t) = 26.67 - \frac{0.8}{0.03} e^{-0.03t}\\$

So, the mass of the salt after t min is

$m(t) = 26.67 - 26.67e^{-0.03t}$

b. When will the concentration of salt in the tank reach 0.1 kg/L?

When the concentration of the salt reaches 0.1 kg/L, m(t) = 0.1 kg/L

Solving the equation for t,

$m(t) = 26.67 - 6.67e^{-0.03t}\\0.1 = 26.67 - 26.67e^{-0.03t}\\26.67e^{-0.03t} = 26.67 - 0.1\\26.67e^{-0.03t} = 26.57\\e^{-0.03t} = 26.56/26.67\\e^{-0.03t} = 0.9963\\$

taking natural logarithm of both sides, we have

-0.03t = ㏑0.9963

-0.03t = -0.0038

t = -0.0038/-0.03

t = 0.124 min

t = 0.124 × 60 s

t = 7.44 s

5 0

3 years ago

Solve: 3x=12 1. 36 2. 4 3. 5 4. 3

Alecsey [184]

Answer:

The answer is 4.

Step-by-step explanation:

This is a basic one-step algebra problem.

We have: 3x = 12.

We would like find what x equals. In order to do that, we must have the x variable on one side of the equals sign, and the rest of the numbers on the other side. In other words, we have to isolate the x variable away from its coefficient.

To do this, we must divide BOTH sides of the equation by 3. This would mean that the 3 on the left side of the equals sign cancels out, totally getting rid of the 3.

Here's what it should look like: $\frac{3x}{3} = \frac{12}{3}$

Now, we actually divide both sides now, which should look like this: $x = 4$

This is because 3 divided by 3 is 1, and 12 divided by 3 is 4.

Hence, our solution for x is 4. Hope this helped!

6 0

2 years ago

Read 2 more answers