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Alborosie
3 years ago
12

PLEASE HELP ASAP 50 POINTS!!!!!

Mathematics
2 answers:
lana [24]3 years ago
8 0

Answer:

the answer is a because all the other ones are multiplying a different number

Stels [109]3 years ago
6 0

Answer:

the answer is a because you have to put the x after the five.

Step-by-step explanation:

I did this last year in my math class

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If the angle of elevation of the sun is 63° and the flagpole casts 27-foot shadow, how tall is the flagpole?
VLD [36.1K]

Answer:

Step-by-step explanation:

To help you solve

Use the fact that (height of flagpole)/(length of shadow) = tan(angle of elevation).

5 0
2 years ago
You are installing new carpeting in a family room. The room is rectangular with dimensions 20 1/2 feet × 13 1/8 feet. You intend
disa [49]

Answer: 1. When you estimate, it is not an exact measurement. 3ft 8 in gets rounded to 4ft and 12 ft 3 in rounds to 12ft. now find the perimeter. P=2l+2w P= 2*12 +2*4 P=32feet

2. 3ft 8in = 3 8/12 or reduced to 3 2/3  12ft 3in = 12 3/12 or reduced to 12 1/4   The fractional part is referring to a fraction of a foot.

3. The perimeter of the room is P=2l+2w or P=2(12 1/4) + 2(3 2/3) p=24 1/2 + 7 1/3  P= 31 5/6 feet

4. The estimate and the actual are very close. They are 1/6 of a foot apart.

5a. Total baseboard 31 5/6ft - 2 1/4 ft = 29 7/12 feet needed.

5b. Take the total and divide it by 8ft  = 29 7/12 divided by 8= 3.7  You are not buying a fraction of a board so you would need 4 boards.

4 0
3 years ago
Please answer all of them! SHOW WORK!!
antiseptic1488 [7]
Wait I can help fm dm
8 0
3 years ago
Express in y=mx+b<br> 3x-2y=8
wel
When rearranged in the format “y = mx + b,” the correct equation would be 2y = -3x - 8.
6 0
3 years ago
41. Two antenna towers, 400 m apart, are to be supported by a single cable anchored at a common point between them. One tower is
Ghella [55]

Answer:

The anchor should be located at the midpoint between the 20m high and 60m high antennas.

Step-by-step explanation:

Let the length of cable for 20m high antenna be represented by x, and that for 60m high antenna be y.

The single length of cable required = x + y.

From the principle of geometry, if the cable is anchored at 200m from the 20m high antenna, it forms a right angled triangle. Applying the Pythagoras theorem,

x = \sqrt{200^{2} - 20^{2} }

 = 199

Applying the same principle to the 60m high antenna gives,

y = \sqrt{200^{2} - 60^{2} }

 = 191

The single length of cable required = 199+  191

                                                           = 390m

Varying the point of location of the anchor between the two antennas causes an increase in the length of cable required.

The anchor should be located at the midpoint between the two antennas to achieve a minimum amount of cable.

5 0
4 years ago
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