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2 years ago

(10^7)^2 as a single exponent

Mathematics

1 answer:

ycow [4]2 years ago

4 0

(10^7)^2
The answer = 10^14

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Solve the following system of equations -8x -8y=24 2x + y=-7

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Solve the equation using substitution?

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3 years ago

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An artificial soccer pitch is 91 meters long and 43 meters wide. If the cost of the pitch is AED 20 per square meter. What is th

Novosadov [1.4K]

The Cost of installing the artificial soccer pitch is $5360

A rectangle is a quadrilateral (has four sides and four angles) in which opposite sides are parallel and equal.

The soccer pitch us rectangle in shape. Hence:

Perimeter of soccer pitch = 2(length + width) = 2(91 + 43) = 268 meters

Cost of installing the pitch = $20 per meter * 268 meters = $5360

The Cost of installing the artificial soccer pitch is $5360

Find out more on perimeter at: brainly.com/question/16596982

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1 year ago

If AB≅DE, BC = EF, and ∠ACB ≅ ∠DFE, then ΔABC ≅ ΔDEF.

Ede4ka [16]

Answer:

false

Step-by-step explanation:

3 0

3 years ago

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Please answer fast<br><br> Thank you!

ziro4ka [17]

Answer:

see explanation

Step-by-step explanation:

the perimeter (P) is the sum of the 4 sides

(a)

P = a + b + c + d

= x + 7 + 2x + 2 + 8x + x + 10 ← collect like terms

= 12x + 19

(b)

If x = 4 , then

P = 12x + 19 = 12(4) + 19 = 48 + 19 = 67

(c)

If x = 2 , then

P = 12x + 19 = 12(2) + 19 = 24 + 19 = 43

(d)

P = a + b + c + d

= x² + 4x + 8 + 2x² + x ← collect like terms

= 3x² + 5x + 8

3 0

9 months ago

What is the exact area and arc length of these sectors? Please helppp

Karolina [17]

<u>Answers with step-by-step explanation:</u>

1. Area of sector 1 = $\frac{90}{360} \times \pi \times 12^2 = 36\pi$

2. Area of sector 2 = $\frac{45}{360} \times \pi \times 19^2 = \frac{2527}{8} \pi$

3. Area of sector 3 = $\frac{270}{360} \times \pi \times 15^2 = \frac{675}{4} \pi$

4. Area of sector 4 = $\frac{270}{360} \times \pi \times 6^2 = 27 \pi$

5. Arc length of sector 1 = $\frac{90}{360} \times 2 \times \pi \times 12 = 6\pi$

6. Arc length of sector 2 = $\frac{315}{360} \times 2 \times \pi \times 19 = \frac{133}{4} \pi$

7. Arc length of sector 3 = $\frac{270}{360} \times 2 \times \pi \times 15 = \frac{45}{2}\pi$

8. Arc length of sector 4 = $\frac{270}{360} \times 2 \times \pi \times 6 = 9\pi$

4 0

2 years ago