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Anna35 [415]
3 years ago
11

Which pair of triangles can be used to show that the slope of line a is the same anywhere along the line?

Mathematics
1 answer:
aalyn [17]3 years ago
7 0

Answer:

C

Step-by-step explanation:

The pair of triangles in option C (image attached below) shows that the slope of line a is the same anywhere on the line.

Let's check it out:

Slope = rise/run

✔️For the first lower triangle:

Rise = 3.5

Run = 4.5

Slope = 3.5/4.5 ≈ 0.8

✔️For the second upper traingle:

Rise = 4

Run = 5

Slope = 4/5 = 0.8

Therefore, it shows that the slope of the line is the same anywhere along the line of the graph.

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Answer:

(-6,4)

Step-by-step explanation:

You started at (-4,2)  The first number in the ordered pair moves the number left and right and the second number moved the point up and down.

We are first told to move the point 2 units to the left.  -4 is my left right number.  If I am at -4 and I go to unites to the left, I will be at -6.  My new point is now (-6,2).  Next we are told to go up 2.  The 2 number in my ordered pair tells me that I am 2 above the x axis.  Now I am going to go two more units up.  I am now at 4, so my new ordered pair after the translation is (-6,4)

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2 years ago
On a map, 1 inch equals 25 miles. Two cities are 2 inches apart on the map. What is the actual distance between the cities?
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The actual distance between the cities is 50 miles
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Marty used the graph to determine if he should go to the beach on vacation. He did not want to go if there were more than 500 pe
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Can someone please help me with these math problems ASAP?!<br> Thanks in advance! ;)
Mamont248 [21]

Answer:

11) Here given Function,

f(x) = \sqrt{2x-11}

And, g(x) = -|0.1|x+5

For f(x) = g(x)

\sqrt{2x-11}=-|0.1|x+5

2x-11=(-|0.1|x+5)^2

2x-11=0.01x^2 - 10|0.1|x+25

2x=0.01x^2 - 10|0.1|x+25+11

2x=0.01x^2 - 10|0.1|x+36

0.01x^2 - 10|0.1|x - 2x+36=0

When we solve this equation,

We found,

 x = 12.5227 ≈ 12.53

Thus, the required solution is, x = 12.53

12) Here the height of rocket A in x second,

f(x) = -16x^2+74x+9

And, The height gain by the rocket B in x seconds,

g(x) = -16x^2+82x

If at x seconds both A and B gain the same height,

That is, f(x) = g(x)

⇒ -16x^2+74x+9= -16x^2+82x

⇒74x + 9 = 82x

⇒ 82x - 74x = 9

⇒ 8x = 9

⇒ x = 1.125 ≈ 1.13

Thus, the required solution is x = 1.13 seconds (approx)


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