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algol13
3 years ago
6

The measures of th eangles in a triangle are in the extended ration 2:5:9. find the angle measures of the triangle.

Mathematics
1 answer:
GarryVolchara [31]3 years ago
6 0
Suppose the angles measures are 2x, 5x, and 9x
the three angles make a total of 180: 2x+5x+9x=180
16x=180
x=11.25
so the three angles are 22.5, 56.25, and 101.25 respectively
a little odd, but those should be the measurements.
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What is the area of the shaded portion of the circle? (16π – 32) in2 (16π – 8) in2 (64π – 32) in2 (64π – 8) in2
GrogVix [38]

Answer:

(16π - 32) in²

Step-by-step explanation:

In the picture attached, the circle is shown.

Area of one-quarter of circle = 1/4*Area of a circle = 1/4*π*radius² = 1/4*π*8² = 16π in²

Area of the triangle = 1/2*base*height = 1/2*8*8 = 32 in²

Shaded area = Area of one-quarter of circle - Area of the triangle = (16π - 32) in²

5 0
3 years ago
What is 70/250 written as a percentage? A) 2.8% B) 28% C) 280% D) 2800%
cluponka [151]
B) 28 percent because it would be 2.8 if you had 7/250
5 0
3 years ago
What is the area of a triangle whose vertices are R(−4, 2) , S(1, 2) , and T(−5, −4) ?
igor_vitrenko [27]
I will solve it using vectors
vector(SR)=(-5,0)
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3 0
3 years ago
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iragen [17]
Answer: I believe it’s D
7 0
3 years ago
In a parallel circuit, ET = 208 V, R = 33 kΩ, and XL = 82 kΩ. What is the apparent power?
jeka94

Answer:

Apparent power is 1.413Volt-Amps.

Step-by-step explanation:

We are given that in a parallel circuit,E_{T}=208V,R=33kΩ and X_{L}=82kΩ.

And we are asked to find apparent power.

Apparent power is the combination of true power and reactive power.

In other words, we can say that in a power triangle apparent power is the hypotenuse where as true power and reactive powers are sides.

Let us find true power , reactive power first and the using Pythagorean theorem we can find apparent power.

True power=\frac{E_{T}^{2}}{R}=\frac{208^{2}}{33000}=1.311watts

Reactive power=\frac{E_{T}^{2}}{X_{L}}=\frac{208^{2}}{82000}=0.528VARs

Hence apparent power=\sqrt{1.311^{2}+0.528^{2}} =1.413VA

5 0
4 years ago
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