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d1i1m1o1n [39]
3 years ago
15

The force F of attraction between two bodies varies jointly as the weights of the two bodies and inversely as the square of the

distance between them. Express this fact as a variation using c as a constant. Use M1 and M2 for the weights of the two bodies.
Mathematics
2 answers:
Gekata [30.6K]3 years ago
4 0
We can write this as

F = c  M1 M2 / d^2    where d = distance between the bodies
zubka84 [21]3 years ago
4 0

Answer:

I think the answer is c=F(d^2)/(m_1m_2)

Step-by-step explanation:

REMEMBER: The formula for joint variation is:

c=y/xz OR y=cxz.

Substitute:

F=Km_1m_2/d^2

then change to:

k= constant, they want you to use "c" instead:

c=F(d^2)/(m_1m_2)

Hope this helps!

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Which of the following describes the graph of y = x² - 225?
lara [203]

Answer:

option 4. The graph has zeroes at x = 15 an x = -15 and it opens upward.

Step-by-step explanation:

we have

y=x^{2} -225

This is the equation of a vertical parabola open upward (the leading coefficient is positive)

The vertex represent a minimum

The vertex is the point (0,-225)

The axis of symmetry is x=0 (y-axis)

Find the x-intercepts (values of x when the value of y is equal to zero)

For y=0

0=x^{2} -225\\x^2=225

square root both sides

x=\pm15

therefore

The graph has zeroes at x = 15 an x = -15 and it opens upward.

3 0
3 years ago
True or False? A circle could be circumscribed about the quadrilateral below.
SVEN [57.7K]

Answer:

<em>The correct answer is:   False</em>

Step-by-step explanation:

<u>If the sum of the opposite angles in a quadrilateral is 180°</u>, then a circle can be circumscribed about the quadrilateral.

Here,  \angle B+\angle D= 110\°+90\° = 180\°

but,  \angle A+\angle C= 70\°+90\°= 160\°

So, a circle can't be circumscribed about the given quadrilateral.

7 0
3 years ago
Read 2 more answers
Find the area and the circumference of a circle with radius 4 yd
Romashka [77]

First, let's find the area! The formula to find the area of a circle is πr²

πr²

3.14 · 4²

50.24 yds²

Now let's find the circumference. The formula for this is 2πr

2πr

2 · 3.14 · 4

25.12 yds

Hope this helps!

4 0
2 years ago
Gabriel wants to paint the model. How
enyata [817]

Answer:

<h2>The area of the base is 144 square inches.</h2><h2>The area of each triangular face is 66 square inches.</h2><h2>Grabiel needs 408 square inches of paint.</h2>

Step-by-step explanation:

The complete problem is attached.

Notice that the figure is a square pyramid, where its base dimensions are 12 inches by 12 inches, which represents an area of

B=12 \times 12 = 144 \ in^{2}

The slant height of the pyramid is 11 inches, which allow us to find the area of each triangle face

A=\frac{1}{2}bh =\frac{1}{2}(12)(11)= 66 \ in^{2}

But there are four triangle faces, so A_{faces} =4(66)=264 \ in^{2}.

Therefore, the area of each triangular face is 66 square inches.

So, the total surface area would be the sum

S=144+264=408 \ in^{2}

Therefore, Gabriel needs 408 square inches to paint the whole model.

4 0
3 years ago
Write the equations for the graphs below (worth 5 points) ​
serg [7]
Answers:

10) y= 1/2x - 2
11) y= 2x + 3
12) y= 2/3x - 4

I found this by using y=mx+ b
3 0
2 years ago
Read 2 more answers
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