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Vadim26 [7]
3 years ago
12

The radius of a semicircle is 5 yards. What is the semicircle's diameter?

Mathematics
1 answer:
mart [117]3 years ago
7 0

Answer:

Diameter: 10 Yards

Step-by-step explanation:

d = 2r

d = 2 ( 5 )

d = 10

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Figure ABCD is a rhombus where the m∠ABC = 84 and m∠ABE = 3x − 6. Solve for x. Rhombus ABCD with diagonals AC and BD and point E
bazaltina [42]

Answer:

<u> x = 16</u>

Step-by-step explanation:

See the attched figure

We should know that, one of the properties of the rhombus is the diagonals bisect the angles of the rhombus.

Given:

m∠ABC = 84  and m∠ABE = 3x − 6

So, m∠ABE = 0.5 * m∠ABC = 0.5 * 84 = 42

∴ 3x - 6 = 42

Solve for x

3x = 42 + 6 = 48

x = 48/3 = 16

<u>∴ x = 16</u>

4 0
4 years ago
Read 2 more answers
Write -2 + 4 + -8 + ...+ 64 +-128 in sigma notation. Then determine the sum.
Grace [21]
It would be -70. Your welcome
4 0
3 years ago
The area of a rectangle is 5/6 sq foot. The width is 2/3 foot. The length of the rectangle is how many feet
max2010maxim [7]

Answer:

The answer is

\frac{5}{4}  \\

Step-by-step explanation:

Area of a rectangle = length × width

Since we are finding the length

<h3>length =  \frac{area}{width}  \\</h3>

From the question

area = 5/6 sq. feet

width = 2/3 foot

Substitute the values into the above formula and solve for the length

The length is

length =  \frac{5}{6}  \div  \frac{2}{3}  \\  =  \frac{5}{6}  \times  \frac{3}{2}  \\  =  \frac{5}{2}  \times  \frac{1}{2}  \\  =  \frac{5}{4}

We have the final answer as

\frac{5}{4}  \\

Hope this helps you

7 0
3 years ago
Read 2 more answers
What are the lengths of the legs of a right triangle in which one acute angle measures 19° and the hypotenuse is 15 units long?
Anarel [89]
The side opposite the angle 19 degrees is given by 15sin 19 = 4.9 units, while the side adjacent the angle 19 degrees is given by 15cos 19 = 14.2 units.
3 0
4 years ago
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Eliminate the parameter.<br><br><br>x = 5t, y = t + 8 (2 points)
maxonik [38]

Answer: y = \frac{1}{5}x + 8

Step-by-step explanation:

       This is asking us to remove the parameter. In other words, we want an equation with only the relation between x and y, so we need to remove the t. There are a few ways to do this, but I am going to set one equation equal to t and then plug it into the next one.

     Given:

x = 5t

    Divide both sides of the equation by 5:

t = \frac{1}{5}x

     -

    Given:

y = t + 8

    Plug in:

y = (\frac{1}{5}x) + 8

    Answer:

y = \frac{1}{5}x + 8

8 0
2 years ago
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