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ohaa [14]
3 years ago
15

Midpoint of (16,5) and (-6,-9)?

Mathematics
1 answer:
Zigmanuir [339]3 years ago
8 0

To solve this problem, we need to use the midpoint formula, where M = (x1+x2/2, y1+y2/2). To solve, we must plug in the given (x,y) values from our ordered pairs and then simplify, shown below:

(x1+x2/2, y1+y2/2)

( (16 + -6)/2, (5 + -9)/2 )

Now, we can begin to simplify by computing the addition in the numerators of both fractions.

(10/2, -4/2)

Next, we can finish the simplification process by dividing these fractions.

(5, -2)

Therefore, the midpoint of (16,5) and (-6,-9) is (5,-2).

Hope this helps!

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According to the Rational Root Theorem, which statement about f(x) = 12x3 – 5x2 + 6x + 9 is true?
Dmitriy789 [7]

The complete statement is (d) Any rational root of f(x) is a factor of 9 divided by a factor of 12.

<h3>How to determine the rational roots?</h3>

The polynomial is given as:

12x^3 - 5x^2 + 6x + 9

The first term in the above equation is 12 i.e. 12x^3, while the last term is 9

To determine the possible rational roots, we divide the factors of the last term i.e. 12 by the factors of the first term i.e. 12

Hence, the complete statement is (d) Any rational root of f(x) is a factor of 9 divided by a factor of 12.

Read more about rational roots at:

brainly.com/question/10937559

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2 years ago
Solve for x: 7 over 8 minus 1 over x equals 3 over 4.
ziro4ka [17]
Im 90% sure its                    x = 8
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5 0
4 years ago
33. Suppose you were on a planet where the
tatyana61 [14]

<u>Answer:</u>

a) 3.675 m  

b) 3.67m

<u>Explanation:</u>

We are given acceleration due to gravity on earth =9.8ms^-2

And on planet given = 2.0ms^-2

A) <u>Since the maximum</u><u> jump height</u><u> is given by the formula  </u>

\mathrm{H}=\frac{\left(\mathrm{v} 0^{2} \times \sin 2 \emptyset\right)}{2 \mathrm{g}}

Where H = max jump height,  

v0 = velocity of jump,  

Ø = angle of jump and  

g = acceleration due to gravity

Considering velocity and angle in both cases  

\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 2}{\mathrm{g} 1}

Where H1 = jump height on given planet,

H2 = jump height on earth = 0.75m (given)  

g1 = 2.0ms^-2 and  

g2 = 9.8ms^-2

Substituting these values we get H1 = 3.675m which is the required answer

B)<u> Formula to </u><u>find height</u><u> of ball thrown is given by  </u>

 \mathrm{h}=(\mathrm{v} 0 * \mathrm{t})+\frac{\mathrm{a} *\left(t^{2}\right)}{2}

which is due to projectile motion of ball  

Now h = max height,

v0 = initial velocity = 0,

t = time of motion,  

a = acceleration = g = acceleration due to gravity

Considering t = same on both places we can write  

\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 1}{\mathrm{g} 2}

where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations

substituting h2 = 18m, g1 = 2.0ms^-2  and g2 = 9.8ms^-2

We get h1 = 3.67m which is the required height

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Step-by-step explanation:

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Answer:

I think the answers is B

Step-by-step explanation:

But im so sorry if its not right...

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