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Elena L [17]
2 years ago
5

The sum of the angles x and y are 127 degrees. If the measure of x is 34 more than half the measure of y, what is the measure of

each angle?
Mathematics
1 answer:
Pepsi [2]2 years ago
8 0
The sum of the angles x and y are 127 degrees.
X + Y = 127

The measure of x is 34 more than half the y.
X = (y/2) + 34
X = (y + 68) / 2

X = 127 - Y

(Y + 68)/2 = 127-Y
254-2Y = Y+68
186 = 3Y
Y = 62
X = 127-62 = 65

The measure of two angles are 62 and 65
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Area addition and subtraction
telo118 [61]

Answer:Area of the shaded region is 73.6 cm^2

Step-by-step explanation:

The circle is divided into two sectors. The Smaller sector contains the triangle. The angle that the smaller sector subtends at the center of the circle is 80 degrees. Since the total angle at the center of the circle is 360 degrees, it means that the angle that the larger sector subtends at the center would be 360 - 80 = 280 degrees

Area of a sector is expressed as

Area of sector = #/360 × πr^2

# = 280

r = 5 cm

Area of sector = 280/360 × 3.14 × 5^2

Area of sector = 61.06 cm^2

Area of the triangle is expressed as

1/2bh = 1/2 × 5 × 5 = 12.5

Area of the shaded region = 61.06 +

12.5 = 73.6

7 0
3 years ago
Which of the following questions describes the equation a - 5 = -12?
Annette [7]
What number, when five is subtracted from it equals twelve
4 0
2 years ago
How many square inches of tin are needed to make the top of a can that is 10 inches in diameter? A) 62.8 in2 B) 78.5 in2 C) 227
Xelga [282]
Area of a circle = PI*radius^2
10 inch diameter = 5 inch radius
Area = PI* 5^2
Area = <span> <span> <span> 78.5398163397 </span> </span> </span>
Answer is B)


8 0
3 years ago
Read 2 more answers
help help help help help help help help help help help help help help help help help help help help help help help help help hel
wel

the answer its -3/2

8 0
3 years ago
Read 2 more answers
(1 ÷ x- 1)+(2÷ x+2)=(3÷2) solve and check for extraneous solutions
Sliva [168]
\frac{1}{x-1}+ \frac{2}{x+2}= \frac{3}{2}
\frac{1(x+2)}{(x-1)(x+2)}+ \frac{2(x-1)}{(x+2)(x-1)}= \frac{3(x-1)(x+2)}{2}
\frac{x+2+2(x-1)}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}
\frac{x+2+2x-1}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}
\frac{3x+1}{(x-1)(x+2)}= \frac{3(x^2+x-2)}{2}
\frac{}{(x-1)(x+2)}= \frac{3x^2+3x-6)}{2}
\frac{3x+1}{(x^2+x-2)}= \frac{3x^2+3x-6)}{2}
\frac{2(3x+1)}{2(x^2+x-2)}= \frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}
\frac{2(3x+1)}{2(x^2+x-2)}-\frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}=0
\frac{2(3x+1)-3x^2+3x-6}{2(x^2+x-2)}=0

this will be continued
5 0
3 years ago
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