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gregori [183]
4 years ago
12

How many 5-millimeter-long spiders would it take to make a line of spiders 1 meter long?

Mathematics
1 answer:
Bogdan [553]4 years ago
3 0

Answer:

1000

Step-by-step explanation:

10milimeter=1 centimeter

100 centimeters=1 meter

10x100=1000

Let me know if im right :)

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the area of a trapezoid is 36 square inches. The height is 4 inches and one bases twice the length of the other base. What are t
vlada-n [284]
The formula for this is A= \frac{1}{2} ( b_{1} + b_{2} )h.  Filling in accordingly, 36= \frac{1}{2}(x+2x)(4).  This simplifies to 36= \frac{1}{2}(3x)(4) and 36=6x.  Therefore, x = 6 so the bases are 6 and 12.
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3 years ago
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I need the answers someone helppppp..
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A. 6 because 30/5= 6
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8×(30+2)=(8×_)+(8×2)
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Using the distributive property, a(b+ c) = ab + ac. In this case, b is 30, so the blank is 30.
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4 years ago
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5(t - 3) -2t = -30 what is the answer?
maxonik [38]

Answer:

t = -5

Step-by-step explanation:

Solve for t:

5 (t - 3) - 2 t = -30

Hint: | Distribute 5 over t - 3.

5 (t - 3) = 5 t - 15:

5 t - 15 - 2 t = -30

Hint: | Group like terms in 5 t - 2 t - 15.

Grouping like terms, 5 t - 2 t - 15 = (5 t - 2 t) - 15:

(5 t - 2 t) - 15 = -30

Hint: | Combine like terms in 5 t - 2 t.

5 t - 2 t = 3 t:

3 t - 15 = -30

Hint: | Isolate terms with t to the left hand side.

Add 15 to both sides:

3 t + (15 - 15) = 15 - 30

Hint: | Look for the difference of two identical terms.

15 - 15 = 0:

3 t = 15 - 30

Hint: | Evaluate 15 - 30.

15 - 30 = -15:

3 t = -15

Hint: | Divide both sides by a constant to simplify the equation.

Divide both sides of 3 t = -15 by 3:

(3 t)/3 = (-15)/3

Hint: | Any nonzero number divided by itself is one.

3/3 = 1:

t = (-15)/3

Hint: | Reduce (-15)/3 to lowest terms. Start by finding the GCD of -15 and 3.

The gcd of -15 and 3 is 3, so (-15)/3 = (3 (-5))/(3×1) = 3/3×-5 = -5:

Answer: t = -5

5 0
4 years ago
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What is the distance between the point (10,12) and (-4,-36)
dimulka [17.4K]
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