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AnnZ [28]
3 years ago
9

Please help! A bicyclist traveled at a constant speed during a timed practice period. Write a proportion to find the distance th

e cyclist traveled in 30 minutes.
Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
3 0
The Answer Would Be x=30
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The sum of twi intergers is 54 and thier difference is 10. Find the integers
JulsSmile [24]

Let x be one number and y be the other.

x + y = 54 ------------- (1)

x - y = 10 ------------- (2)

.

<u>Add (1) and (2):</u>

2x = 64

<u>Divide both sides by 2:</u>

x = 32

-

<u>Plug x = 32 into (1):</u>

32 + y = 54

<u>Take away 32 from both sides:</u>

y = 22

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Answer: The two numbers are 32 and 22.

4 0
3 years ago
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How do you write 715.7 in scientific notation
My name is Ann [436]

Answer:

<em>We </em><em>can </em><em>write </em><em>7</em><em>1</em><em>5</em><em>.</em><em>7</em><em> </em><em>in </em><em>scientific </em><em>notation </em><em>by </em>

<em>7</em><em>.</em><em>1</em><em>5</em><em>7</em><em>*</em><em>1</em><em>0</em><em>^</em><em>2</em>

7 0
3 years ago
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Point C is located at (1, 2) and point D is located at (−4, −2). Find the x value for the point that is 1\4 the distance from po
max2010maxim [7]

The easiest way to do this is to realize this is a <em>weighted average</em> of the two points.  (A weighted average is a linear combination where the coefficients add to one.)


(x,y) = (1-t)C + tD


where t is a real parameter. When t=0 we're at point C, when t=1 we're at point D. We're interested in t=1/4,


(x,y)= (3/4) C + (1/4) D


We're only asked for the x coordinate:


x = (3/4) (1) + (1/4) (-4) = 3/4 - 1 = - 1/4


Answer: \quad - \dfrac 1 4



7 0
3 years ago
Find cos 0, where 0 is the angle shown.<br> Give an exact value, not a decimal approximation.
marin [14]

Answer:

cosθ = \frac{5}{13}

Step-by-step explanation:

This is a 5- 12- 13 right triangle then

cosθ = \frac{adjacent}{hypotenuse} = \frac{5}{13}

5 0
3 years ago
Please I need help the answer please
Volgvan

Answer:-72

Step-by-step explanation:

(6)*(-12)

-72

5 0
3 years ago
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