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Nastasia [14]
2 years ago
6

Jean and Jericho who are playing in the school grounds decided to sit on a

Mathematics
1 answer:
Natalija [7]2 years ago
7 0

We want to see underline the correct part in each statement.

  • 1) This situation illustrates (direct, <u>inverse</u>) variation.
  • 2) The two quantities that must vary in this situation are (<u>weight and </u>
  • <u>distance from the cente</u><u>r</u>, height and weight).
  • 3) The heavier the kid, the (<u>closer</u>, farther) he/she should be at the
  • center to balance the seesaw?
  • 4) When Jean moves farther from the center, Jericho tends to go (up,
  • <u>down</u>).
  • 5) If Jericho moves closer to the center, Jean tends to go (up, <u>down</u>).

So, Jean and Jericho are playing on a seesaw.

Jean is heavier than Jericho.

Now, notice that a seesaw is a lever. So it "amplificates" the force that you apply in one end to lift the weight that is on the other end. Depending on the form of the lever and the weights, the force that you need to do changes.

If we define:

  • W₁ = Jean's weight.
  • d₁ = Jean's distance to the center.
  • W₂ = Jericho's weight.
  • d₂ = Jericho's distance to the center.

We must have:

W₁*d₁ = W₂*d₂

Then:

1) This situation illustrates (direct, <u>inverse</u>) variation.

d₁, the position of Jean, varies inversely with respect to Jean's weight.

2) The two quantities that must vary in this situation are (<u>weight and </u>

<u>distance from the center</u>, height and weight).

(by the equation above)

3) The heavier the kid, the (<u>closer</u>, farther) he/she should be at the

center to balance the seesaw?

By the given equation, we see that d₁ must be smaller than d₂.

And the last two are trivial:

4) When Jean moves farther from the center, Jericho tends to go (up,

<u>down</u>).

5) If Jericho moves closer to the center, Jean tends to go (up, <u>down</u>).

If you want to learn more, you can read:

brainly.com/question/18320907

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Please answer this question now
Alexeev081 [22]

Answer:

Area = 400.4 m^2

Step-by-step Explanation:

Given:

∆UVW,

m < U = 33°

m < V = 113°

VW = u = 29 m

Required:

Area of ∆UVW

Solution:

Find side length UV using Law of Sines

\frac{u}{sin(U)} = \frac{w}{sin(W)}

U = 33°

u = VW = 29 m

W = 180 - (33+113) = 34°

w = UV = ?

\frac{29}{sin(33)} = \frac{w}{sin(34)}

Cross multiply

29*sin(34) = w*sin(33)

Divide both sides by sin(33) to make w the subject of formula

\frac{29*sin(34)}{sin(33)} = \frac{w*sin(33)}{sin(33)}

\frac{29*sin(34)}{sin(33)} = w

29.77 = w

UV = w = 30 m (rounded to nearest whole number)

Find the area of ∆UVW using the formula,

area = \frac{1}{2}*u*w*sin(V)

= \frac{1}{2}*29*30*sin(113)

= \frac{29*30*sin(113)}{2}

Area = 400.4 m^2 (to nearest tenth).

4 0
3 years ago
Help plz idk okhehehehehe​
TiliK225 [7]
That is the answer for the first one

6 0
2 years ago
Savings are what percentage of monthly net income? Round your answer to the nearest whole percent.
cricket20 [7]
To find the percent of the monthly income that is savings, you will divide the amount saved by the total amount of income.

300/1387 = 0.216

The approximate percentage of income that is savings is 22%.

7 0
3 years ago
Kennedy and Kenyon are meeting at Dave and Buster’s to play video games after school one day. They arranged to meet at 4:45. Ken
melisa1 [442]

Answer:

5:00 o clock

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Verifying Trig Functions
anzhelika [568]

Answer:

The answer to your question is below

Step-by-step explanation:

                csc α - cot α = \frac{sin \alpha }{1 + cos \alpha }

Work with the left part

                \frac{1}{sin\alpha } - \frac{cos \alpha }{sin \alpha}

Simplify

                \frac{1 - cos\alpha }{sin\alpha }

Multiply by the reciprocal

 \frac{1 - cos\alpha }{sin \alpha } x \frac{1 + cos\alpha }{1 + cos\alpha }                                  

Simplify

\frac{1 - cos^{2}\alpha}{sin\alpha (1 + cos\alpha)}

Remember that

                 sin²α = 1 -cos²α

Substitute the previous equation

\frac{sin^{2}\alpha }{sin\alpha(1 + cos\alpha ) }

Simplify

\frac{sin\alpha}{1 + cos \alpha}      

The identity i proved.            

5 0
3 years ago
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