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QveST [7]
4 years ago
14

Find the length of the radius of the circle.

Mathematics
2 answers:
dimaraw [331]4 years ago
8 0

Answer:

C. 6 inches

Step-by-step explanation:

Because the triangle is a right triangle you can use the Pythagorean Theorem. Use the formula

x2 + x2 =   (72)

2 and solve for x. The equation simplifies to 2x2 = 72; x2 = 36 so x = 6.

Ymorist [56]4 years ago
3 0

Answer:

C. 6 inches

Step-by-step explanation:

It is a 45, 45, 90 triangle because there is a 90 degree angle and since both of the legs are equal it is an isosceles triangle so the two angles are equal, so they can both only be 45 because together they have to equal 90.

That means the two radii are x and the hypotenuse is x√2   so √72 = x√2

√72 = x√2

now you divide both sides by √2 so that the x is alone

√72/√2 = x                you have to multiply the √72/√2  by √2   because the denominator can not be a radical, You dont have to do it to the x because you are multiply both the numerator and denominator by√2 so it is like 1

√72/√2  *  √2/√2

√144/√4            the √144 is 12 and the √4 is 2

12/2            now simplify

6

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Answer:

Step-by-step explanation:

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Answer: An inequality

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An inequality is a mathematical sentence written with a greater than

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3 years ago
Read 2 more answers
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

6 0
4 years ago
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