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Lady bird [3.3K]
3 years ago
8

How can you express (32 + 56) as a multiple of a sum of whole numbers with no common factor?

Mathematics
1 answer:
bazaltina [42]3 years ago
5 0
To express (32+56) as a multiple sum of whole number we proceed as follows;
32=2*2*2*2*2=2^5
56=2*2*2*5=(2^3)*5
Thus;
(32+56)
=2^5+2^3*5
=2^3(2^2+5)
The answer is:
2^3(2^2+5)

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Triangle ABC is inscribed in the circle, mA=50, and AB=BC. Find arch AC.
Karolina [17]

The answer is C) 160.

We know this because if mA = 50, we know that mC must also be 50. This is due to the fact that AB = BC. This leaves us with mB as 80 since the angles of a triangle always have to equal 180.

Now knowing this, it is easy to find the arc lengths in degrees. When you have a transcribed triangle, all we are going to do here is double the angle of the triangle to get the arc measure.

mB = 80

80*2 = 160

3 0
3 years ago
To cover A rectangle region of her yard penny needs atleast 170.5 square feet of sod. The length of the region is 15.5 feet. Wha
stiks02 [169]
The area needed is at least 170.5 ft².

Let w  =  the width of the rectangular area.
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7 0
2 years ago
Convert into percentages 7/25
LekaFEV [45]

Answer:

28%

Step-by-step explanation:

<u>Divide the numerator by the denominator in order to convert this fraction into a decimal.</u>

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8 0
3 years ago
What is the area of this figure? Enter your answer in the box. m² A parallelogram with a right triangle created inside it with a
igomit [66]

Answer:

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6 0
3 years ago
Read 2 more answers
Which sum or difference identity would you use to verify that cos (180° - q) = -cos q?
Phantasy [73]

Answer:

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Step-by-step explanation:

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We have to write which identity we will use to prove the given statement.

Consider \cos (180^{\circ}-q)=-\cos q

Take left hand side of given expression \cos (180^{\circ}-q)

We know

\cos (a-b)=\cos a \cos b+\sin a \sin b

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Substitute , we get,

\cos (180^{\circ}-q)=\cos 180^{\circ}  \cos (q)+\sin q \sin 180^{\circ}

Also, we know \sin 180^{\circ}=0 and \cos 180^{\circ}=-1

Substitute, we get,

\cos (180^{\circ}-q)=-1\cdot \cos (q)+\sin q \cdot 0

Simplify , we get,

\cos (180^{\circ}-q)=-\cos (q)

Hence, use difference identity to  prove the given result.

7 0
3 years ago
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