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UNO [17]
3 years ago
7

Find m measure of angle YQR=? Degrees

Mathematics
1 answer:
Nimfa-mama [501]3 years ago
7 0

Answer:

21°

Step-by-step explanation:

Since these are both equivalent right triangles, ∡RQY and ∡PQY are the same. So set them equal to each other.

16x-27=5x+6

x=3

Next plug x into ∡RQY

16(3)-27=21°

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Your job in a company is to fill quart-size bottles of oil from a full 100-gallon oil tank. Then you are to pack 12 quarts of oi
Andreyy89

Answer:

33 cases of oil

Step-by-step explanation:

You start with 100 gallons of oil.

1 gallon = 4 quarts

100 gallons = 100 * 4 quarts

100 gallons = 400 quarts

You start with 400 quarts.

You place 12 quarts in each box.

400/12 = 33 1/3

You can pack 33 full cases plus 1/3 of another case.

The question only askes about full cases.

Answer: 33 cases of oil

5 0
3 years ago
Which transformation shows a relfection of DEF?
AfilCa [17]

Answer:

The middle option.

Step-by-step explanation:

This is because the points on each triangle are equidistant from the imaginary line in between the two triangles.  It isn't the first option because that is a rotation, and it isn't the third right option because that is a translation.

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6 0
3 years ago
Decide whether or not it is possible for a triangle to have three angle measurements or three side lengths given. If it is possi
shutvik [7]

Answer:

First one is not possible

Second one is possible, and would be an isocèles triangle

Third one is possible, would be scalene

Last one is not possible

Comment any further questions :)

Step-by-step explanation:

5 0
3 years ago
Short Response
garik1379 [7]

Answer:

(a)2(89x+84)

(b)2(89x+84-x^2\pi)

Step-by-step explanation:

The dimensions of the larger rectangular field are:

  • Length=5x + 12; Width = 9x + 14.

The dimensions of the smaller rectangular soccer field are:

  • Length=5x; Width = 9x.

(a)Area of the part of the field that is outside the soccer field

=Area of the larger rectangular field - Area of the Soccer Field

=(5x+12)(9x+14)-5x(9x)

=(5x)(9x)+70x+108x+168-5x(9x)

=178x+168

=2(89x+84)

(b)Radius of the Semicircular Fountain =2x

From Part (a),

Area of the larger rectangular field - Area of the Soccer Field=178x+168

Area of the Semicircular Fountain =\dfrac{\pi r^2}{2} =\dfrac{\pi (2x)^2}{2} =\dfrac{4x^2\pi}{2} =2x^2\pi

Area of the Field that does not include the soccer  field or the fountain.

=Area of the larger rectangular field - Area of the Soccer Field-Area of the Semicircular Fountain

=178x+168-2x^2\pi\\=2(89x+84-x^2\pi)

8 0
3 years ago
Help me plz i need help on math i will give THANKS
atroni [7]
The answer is for your paper is 21
8 0
3 years ago
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