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Nonamiya [84]
3 years ago
7

Find the values of x and y x=42, y=26.3 x=47, y=30 x=42, y=30 x=47, y=26.3

Mathematics
2 answers:
olga55 [171]3 years ago
3 0

Answer:

x = 42, y = 30

Step-by-step explanation:

From the given figure, we have to find the value of x and y.

The measurement of an angle on a straight angle is 180 degrees.

Using this concept, we have

3y+5+2x+1=180\\3y+2x=174....(1)

Now, we know that the sum of consecutive interiors angles is 80 degrees. Hence, we have

3y+5+85=180\\3y+90=180\\3y=90\\y=30

Plugging the value of y in equation (1)

3(30)+2x=174\\90+2x=174\\2x=84\\x=42

Hence, the values of x and y are

x = 42, y = 30

Lunna [17]3 years ago
3 0
2x +1= 85
2x=84
x=42

180-85=95

3y+5=95
3y=90
y=30
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How do you do this question?
alina1380 [7]

Answer:

(8√2) / 15

Step-by-step explanation:

A curve bounded by the y-axis is represented by in terms of dy;

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Our solutions (0, 2) are our limits. The area of the curve is in the form A\:=\:\int _b^a\:f\left(t\right)g'\left(t\right)dt , so now let's introduce the limits of integration, x(t) and dy/dt. Remember, dy/dt = (1/2 * 1/√t) (second equation). 1/2 * 1/√t can be rewritten as 1/2 * t^(-1/2)....

A\:=\:\int _2^0\:\left(t^2-2t\right)\left(\frac{1}{2}t^{-\frac{1}{2}}\right)dt\\\\= \int _2^0\:\left(\frac{1}{2}t^{\frac{3}{2}}-t^{\frac{1}{2}}\right)dt\\\\= \left[\frac{t^{\frac{5}{2}}}{5}-\frac{2t^{\frac{3}{2}}}{3}\right]_2^0\\\\= 0\:-\:\left(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)\\\\= \frac{8\sqrt{2}}{15}

Your solution is 8√2 / 15

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3 years ago
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