Answer:
150
Step-by-step explanation:
The are is the sum of are of square and area of triangle.
A=A1+A2
A1=a*a=10*10=100
A2=a*h/2=10*10/2=50
A=100+50=150
Answer:
3
Step-by-step explanation:
A ball is 10 meters. Since 100 centimeters are in a meter, each ball has 100*10 = 1000 centimeters.
The teacher needs 70 strings each 40 centimeters. The total amount they need is 70*40 = 2800 centimeters.
Since a ball has 1000 centimeters, 2800 centimeters will need 3 balls for a total of 3000 centimeters.
Answer:
81
Step-by-step explanation:
3*3=9
9*3=27
27*3=81
Answer:
9. m(YZ) = 102°
10. m(JKL) = 192°
11. m<GHF = 75°
Step-by-step explanation:
9. First, find the value of x
4x + 3 = 3x + 15 (inscribed angle that are subtended by the same arc are equal based on the inscribed angle theorem)
Collect like terms
4x - 3x = -3 + 15
x = 12
4x + 3 = ½(m(YZ)) (inscribed angle of a circle = ½ the measure of the intercepted arc)
Plug in the value of x
4(12) + 3 = ½(m(YZ))
48 + 3 = ½(m(YZ))
51 = ½(m(YZ))
Multiply both sides by 2
51*2 = m(YZ)
102 = m(YZ)
m(YZ) = 102°
10. First, find the value of x.
7x + 5 + 6x + 6 = 180° (opposite angles in an inscribed quadrilateral are supplementary)
Add like terms
13x + 11 = 180
13x = 180 - 11
13x = 169
x = 169/13
x = 13
7x + 5 = ½(m(JKL)) (inscribed angle of a circle = ½ the measure of the intercepted arc)
Plug in the value of x
7(13) + 5 = ½(m(JKL))
96 = ½(m(JKL))
Multiply both sides by 2
2*96 = m(JKL)
m(JKL) = 192°
11. First, find x.
5x + 15 = ½(11x + 18) (inscribed angle of a circle = ½ the measure of the intercepted arc)
Multiply both sides by 2
2(5x + 15) = 11x + 18
10x + 30 = 11x + 18
Collect like terms
10x - 11x = -30 + 18
-x = -12
Divide both sides by -1
x = 12
m<GHF = 5x + 15
Plug in the value of x
m<GHF = 5(12) + 15
m<GHF = 60 + 15
m<GHF = 75°
