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suter [353]
1 year ago
12

What can be concluded from the statement ∠1 + ∠2 = 90°?

Mathematics
1 answer:
irinina [24]1 year ago
8 0
  • <1+<2=90°

The sum of measures of angles A and B is 90°

Hence the angles are complementary

  • As complementary angles have sum 90°

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This is an isosceles trapezoid.
Ad libitum [116K]

Answer:

105°

Step-by-step explanation:

X and 75 has to add up to 180

so 180 - 75 = 105°

3 0
2 years ago
The circumference of a circle is 36 cm.
dusya [7]
18cm (A). Remember to get the radius when you have the circumference is to ALWAYS divide it by 2. 36/2 = 18
6 0
3 years ago
6. Dina is moving and he is going to rent a moving van. She will need it for one day. The cost of renting the van $145 for the d
Marat540 [252]

Answer:

403.75

Step-by-step explanation:

we can write the following equation

.75m+145=price

we just need to plug in 345 for m

.75*345+145=price

put this into a calculator and get 403.75

6 0
3 years ago
What is 37.5 percent of 80
Leya [2.2K]
37.5*0.010=.375 (or press % button on calculator) 
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7 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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