Ok It takes 46 minutes for 10 people to paint 3 walls. How many minutes does it take 4 people to paint 6 walls

1 answer:

3 0

To solve this problem, all we have to do is to closely analyze the situation and use the principle of ratio and proportion.

What we have to do first is to write the given values in terms or units of job / (man hour). In this case, the job refers to the number of walls that can be painted by the specified number of man and number of hours.

For the 1st case:

3 walls / (10 people * 46 minutes)

For the 2nd case:

6 walls / (4 people * t)

Now to solve for t, we equate the two cases:

3 walls / (10 people * 46 minutes) = 6 walls / (4 people * t)

t = (6 walls / 4 people) / [3 walls / (10 people * 46 minutes)]

t = 1.5 / (3 / 460)

t = 230 minutes

Therefore it requires 230 minutes for 4 people to paint 6 walls.

You might be interested in

How do you graph y= square root 4-x

Ksivusya [100]

Hi! Your answer is below!

Answer- Here's a simple model:

X⟶f⟶Y

This is just something to help you out! I found it while searching on the internet! I have no Idea what the answer is, but in my defense, I think it's going to be choice 3!

Srry if I couldn't help! Glad if I could help!

6 0

3 years ago

Looking at the top of tower A and base of tower B from points C and D, we find that ∠ACD = 60°, ∠ADC = 75° and ∠ADB = 30°. Let t

katrin2010 [14]

Answer:

$\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24$

Step-by-step explanation:

We can use the Law of Sines to find segment AD, which happens to be a leg of $\triangle ACD$ and the hypotenuse of $\triangle ADB$ .

The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

$\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}$

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is $\angle CAD$ . The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

$\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}$

Now use this value in the Law of Sines to find AD:

$\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}$

Recall that $\sin 45^{\circ}=\frac{\sqrt{2}}{2}$ and $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ :

$AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}$

Now that we have the length of AD, we can find the length of AB. The right triangle $\triangle ADB$ is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio $x:x\sqrt{3}:2x$ , where $x$ is the side opposite to the 30 degree angle and $2x$ is the length of the hypotenuse.

Since AD is the hypotenuse, it must represent $2x$ in this ratio and since AB is the side opposite to the 30 degree angle, it must represent $x$ in this ratio (Derive from basic trig for a right triangle and $\sin 30^{\circ}=\frac{1}{2}$ ).