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3 years ago

How do you calculate the circumference of a sphere if you only have thd measurements of the diameter

Mathematics

2 answers:

Setler79 [48]3 years ago

7 0

Hope it helps .

Step-by-step explanation:

Solve the equation for the diameter of the circle, d= C/π. In this example, "d = 12 / 3.14." or "The diameter is equal to twelve divided by 3.14." Divide the circumference by pi to get the answer.

AleksAgata [21]3 years ago

4 0

Answer:

Solve the equation for the diameter of the circle, d= C/π. In this example, "d = 12 / 3.14." or "The diameter is equal to twelve divided by 3.14." Divide the circumference by pi to get the answer. In this case, the diameter would be 3.82 inches.

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PY=2, HP=3, find HY<br><br> Geometry

klemol [59]

Answer:

$HY=\sqrt{5}\ units$

Step-by-step explanation:

we know that

The triangle HPY is a right triangle (because the angle ∠HYP is a right angle)

Applying the Pythagorean Theorem

$HP^2=HY^2+PY^2$

substitute the given values

$3^2=HY^2+2^2$

Solve for HY

$9=HY^2+4$

$HY^2=9-4$

$HY^2=5$

$HY=\sqrt{5}\ units$

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3 years ago

A television that was originally marked $199.95 is now 25% off. Write a linear expression for the new cost of the television. Th

ioda

Answer is c

199.95 x 25/100

=49.9875

Sale price

=199.95-49.9875

=149.9625

=149.96

5 0

3 years ago

Read 2 more answers

Compare the slopes of parallel lines and perpendicular lines?

Gnesinka [82]

Answer: C

Step-by-step explanation: l-7&!:

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3 years ago

Write y=-(3/5)x-2 in standard form using integers <br> please thank you

juin [17]

Y= -3/5x - 2 Add -3/5x to both sides
3/5x +y= -2 Multiply both sides by 5 because 5 is the denominator.
By multiplying by 5 you can cancel out and make it a real number.
5(3/5x +y) = -2(5)

3x + 5y= -10 is the answer.

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3 years ago

Davis Construction is building a new housing development. They begin by mapping the development out on a coordinate grid. They p

Mandarinka [93]

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -4}}\quad ,&{{ -1}})\quad 
% (c,d)
&({{ 6}}\quad ,&{{ 3}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left(\cfrac{{{ 6}} -4}{2}\quad ,\quad \cfrac{3-1}{2} \right)$

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3 years ago