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docker41 [41]
3 years ago
7

Help plzzz... circle geometry question.

Mathematics
1 answer:
Elodia [21]3 years ago
7 0

Answer:

IU = 161

Step-by-step explanation:

Angle Formed by Two Secants=  1/2 (difference of Intercepted Arcs)

42 = 1/2 (7m+5 - (3m-1))

Distribute the minus sign

42 = 1/2 (7m+5 - 3m+1))

Combine like terms

42 = 1/2 ( 4m+6)

Distribute the 1/2

42 = 2m+3

Subtract 3 from each side

42-3 = 2m+3-3

39 = 2m

Divide by 2

19.5 = m

The sum of the arcs = 360

IU+ 3m-1 + 7m+5 = 360

IU +10m+4 = 360

IU +10(19.5) +4 = 360

IU +195+4 = 360

IU = 360 - 199

IU=161

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Pedro is building a playground in the shape of a right triangle. The base of the triangle is 18 yards. The height of the triangl
Bezzdna [24]

Answer:

567 square yards

Step-by-step explanation:

The Area for 1 Triangular playground = 1/2 × Base × Height

The base of the triangle is 18 yards. The height of the triangle is 21 yards.

Hence:

= 1/2 × 18 yards × 21 yards

= 189 square yards

The area of the Triangular 3 playground is given as:

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6 0
3 years ago
39 s 150% of what number?
professor190 [17]

Answer:

26

Hope this helps!

6 0
3 years ago
Are 9x and 9x2 like terms? Explain why or why not?
enot [183]

Answer:

no

Step-by-step explanation:

they are not like terms because the term 9x^2 is to the 2nd power which means that you are supposed to times the number by itself 2 times and 9x is not to the 2nd power so it stays the same.

4 0
3 years ago
Find the number b such that the line y = b divides the region bounded by the curves y = 36x2 and y = 25 into two regions with eq
Gemiola [76]

Answer:

b = 15.75

Step-by-step explanation:

Lets find the interception points of the curves

36 x² = 25

x² = 25/36 = 0.69444

|x| = √(25/36) = 5/6

thus the interception points are 5/6 and -5/6. By evaluating in 0, we can conclude that the curve y=25 is above the other curve and b should be between 0 and 25 (note that 0 is the smallest value of 36 x²).

The area of the bounded region is given by the integral

\int\limits^{5/6}_{-5/6} {(25-36 \, x^2)} \, dx = (25x - 12 \, x^3)\, |_{x=-5/6}^{x=5/6} = 25*5/6 - 12*(5/6)^3 - (25*(-5/6) - 12*(-5/6)^3) = 250/9

The whole region has an area of 250/9. We need b such as the area of the region below the curve y =b and above y=36x^2 is 125/9. The region would be bounded by the points z and -z, for certain z (this is for the symmetry). Also for the symmetry, this region can be splitted into 2 regions with equal area: between -z and 0, and between 0 and z. The area between 0 and z should be 125/18. Note that 36 z² = b, then z = √b/6.

125/18 = \int\limits^{\sqrt{b}/6}_0 {(b - 36 \, x^2)} \, dx = (bx - 12 \, x^3)\, |_{x = 0}^{x=\sqrt{b}/6} = b^{1.5}/6 - b^{1.5}/18 = b^{1.5}/9

125/18 = b^{1.5}/9

b = (62.5²)^{1/3} = 15.75

8 0
3 years ago
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Answer:

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