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Fed [463]
3 years ago
7

a frame for mrs. nunez's class picture has an area of 93.5 square inches. the length of the frame measures 8 1/2 inches. what mu

st be the width of the frame?
Mathematics
1 answer:
kakasveta [241]3 years ago
7 0

Answer:

Step-by-step explanation:

length times width = area

so you have 8.5w=93.5

divide 8.5 on both sides to get your answer

w= 11

so your width must be 11 inches

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(60y – 18) = 2(8-16)​
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3/10

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3 years ago
Zaheer, a boy of height 1.5m was watching the entire programme. Initially he observed the top of the flag pole at an angle of el
7nadin3 [17]

Answer:

15.2 m

Step-by-step explanation:

You need to draw a figure. Start by drawing a horizontal segment approximately 10 cm long; that is the ground. Label the left end point A and the right endpoint B. On the right endpoint, B, go up a short 1 cm vertically. That is 1.5 m, the height of Zaheer. Label that point C. Now from that point draw a horizontal line that ends up above point A. Label that point D. Now go back to point C. Draw a segment up to the left at a 30 deg angle with CD. End the segment vertically above point D. Label that point E. That is the top of the flagpole. Draw a vertical segment down from point E through point D ending at point A. Segment AE is the flagpole. Go back to point C. Move 3 cm to the left on segment CD, and draw a point there and label it F. That is where Zaheer moved to. Now connect point F to point E. That is a 45-deg elevation to point E, the top of the flagpole.

m<EFD = 45 deg

m<EFC = 135 deg

m<FEC = 15 deg

m<ECD = 30 deg

We now use the law of sines to find EC

(sin 15)/10 = (sin 135)/EC

EC = 27.32

Because of the 30-60-90 triangle, ED = EC/2

ED = 13.66

Now we add the height of Zaheer to find AE.

13.66 + 1.5 = 15.16

Answer: 15.2 m

7 0
3 years ago
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