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lina2011 [118]
3 years ago
14

A circle has a radius of 17 mm. An arc length of the circle is 30 mm. What is the area of the sector that is bound by the arc

Mathematics
2 answers:
nikklg [1K]3 years ago
7 0

Answer:

255mm squared

Step-by-step explanation:

it is true i got it wrong and found out the hard way

Mama L [17]3 years ago
7 0

Answer:

255

Step-by-step explanation:

edge

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A company earns a profit of $100 the first month it is in business. Every month after that, the company earns a profit that is 1
ryzh [129]

Answer:

506.25

Step-by-step explanation:

Month 1: $100

Month 2: 100 times 1.5 = 150

Month 3: 150 times 1.5 = 225

Month 4: 225 times 1.5 = 337.50

Month 5: 337.50 times 1.5 = 506.25

6 0
3 years ago
4/6 simplfy it can someone help
Gemiola [76]

4/6=2/3 (divide both numerator and denominator by 2 (gcd)).

Hope this helps.

4 0
3 years ago
Read 2 more answers
I) Construct a triangle PQR such that |PQ|=8cm,{RPQ=90°{PQR=30°.Measure |RQ|​
scoray [572]

Answer:

6.93 cm

Step-by-step explanation:

You have a right triangle (90°), so you do as follow:

If I understand correctly, you are looking for the hypotenuse so

cos(30) = \frac{PQ}{QR} = \frac{8 cm}{QR}

That is equal to QR = cos(30)*8 cm = 6.928 cm

5 0
3 years ago
Can someone please HELP! FOR BRAINLIEST!
RoseWind [281]

Answer:

1st pic:

step 2.

2nd pic:

Rene should add, not multiply 2 volumes.

The correct answer is 48+216 =264

3rd pic:

a, for blue prism: V = Area of base x Height = 48 x a =720

=> a = 720/48 = 15

b,  for orange prism: V = Area of base x Height = (12xb)/2 x 15 =720

=> b =(720x2)/(15x12)= 8

8 0
3 years ago
How much material is needed to construct a triangular tent that is 10 feet wide, 12 feet tall, and 18 feet long with side measur
Ket [755]
The answer is " 768 square feet. ".

Total surface area of a triangular prism = width x height +(sum of sides + width) x length and here;
width = 10
height = 12
sides = 13
length = 18
= (10 x 12) + (13 + 13 + 10) x 18
= 120 + (36 x 18)
= 120 + 648
= 768 square feet. 
3 0
3 years ago
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