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motikmotik
3 years ago
5

Teacher takes about 25 minutes to make out a test for a mathematics class. How long will it take to make out tests for all five

classes?
Mathematics
1 answer:
Akimi4 [234]3 years ago
6 0

Answer:

2h 5min

Step-by-step explanation:

1 class: 25 min

5 classes: 25 x 5 = 125 min = 2h 5min

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Use a net to find the surface Area of the square pyramid The height of the pyramid is 9 yes and the base sides are all 2.5 yrds
Cerrena [4.2K]

Answer:

Area = 51.68yd^2

Step-by-step explanation:

Given

h = 9yd -- height

a = 2.5yd --- base sides

Required

Determine the surface area

The net is not given. So, I will solve directly.

The surface area is calculated as:

Area = a^2 + 2a\sqrt{\frac{a^2}{4} + h^2}

So, we have:

Area = 2.5^2 + 2*2.5\sqrt{\frac{2.5^2}{4} + 9^2}

Area = 6.25 + 5\sqrt{\frac{6.25}{4} + 81}

Area = 6.25 + 5\sqrt{1.5625 + 81}

Area = 6.25 + 5\sqrt{82.5625}

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Area = 51.68yd^2

6 0
3 years ago
Select the decimal that is equivalent to 101/500
s344n2d4d5 [400]

The answer should be .202. If the question is looking for rounding, it might want .2 instead, but only if it tells you to round. Just divide 101 by 500 in a calculator and that's your answer.

Hopefully this helps! Let me know if you have any more questions.

4 0
2 years ago
Read 2 more answers
The center of a circle is at (2, −5) and its radius is 12.
stiv31 [10]

Answer:

The equation of the circle is (x - 2)² + (y + 5)² = 144 ⇒ A

Step-by-step explanation:

The form of the equation of the circle is (x - h)² + (y - k)² = r², where

  • r is the radius of the circle
  • h, k are the coordinates of the center of the circle

Let us solve the question

∵ The center of the circle is at (2, -5)

→ From the rule above

∴ h = 2 and k = -5

∵ The radius of the circle is 12

∴ r = 12

→ Substitute the values of r, h, and k in the form of the equation above

∵ (x - 2)² + (y - -5)² = (12)²

∴ (x - 2)² + (y + 5)² = 144

∴ The equation of the circle is (x - 2)² + (y + 5)² = 144

7 0
3 years ago
A point charge q1 = 3.30nC is located on the x-axis at x= 1.90m , and a second point charge q2= -6.40 nC is on the y-axis at y=
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Answer:

Step-by-step explanation:

charge q1 is placed at x = 1.90 m and the charge q2 is placed at y = 1.15 m

here, the charge enclosed in a sphere is zero as the radius of sphere is 0.625 m which is less than the x = 1.90 m and y = 1.15 m. So by the Gauss's theorem,

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where q is the charge enclosed, as the charge enclosed is zero so the electric flux is zero.

3 0
2 years ago
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sorry

Step-by-step explanation:

i dont see the question i can help if i see the question

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