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Aloiza [94]
3 years ago
11

Complete the following statement. -4(9 + 6) =

Mathematics
1 answer:
Advocard [28]3 years ago
5 0
-60 lovee , you welcome .
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Li believes that the graph shows a direct variation why is li corect​
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Because it directly passes through the x and y axis
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A line segment has more then one midpoint true or false
Anika [276]

Answer:

false

Step-by-step explanation:

any line segment or distance has only one midpoint.

there is only one middle to everything.

how would that look like, if something had 2 middles ?

then they would not be a middle, and the middle between these 2 middles would then be the real middle ...

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a ramp measures 6 feet long. if the ramp is 12 inches tall, what is the horizontal distance it covers? i'll give brainliest to w
luda_lava [24]

we know that

1 ft--------> is equals to 12 in

the ramp is 12 inches tall----------> 1 ft tall

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applying the Pythagorean theorem

c²=a²+b²

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c-----> 6 ft long

a----> horizontal distance

b-----> 1 ft tall

a²=c²-b²------> a²=6²-1²-----> a²=35------> a=√35------> a=5.92 ft

the answer is

5.92 ft

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2 years ago
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Find the average. football yardage: 10 yd, –5 yd, 7 yd, 9 yd, –11 yd
CaHeK987 [17]

Answer:

10

Step-by-step explanation:

Add all of the numbers up and divide by the amount of numbers there are. You subtract for the negatives.

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The points A BC :(2, 2,1), :(1,1,3), :(2,0,5) − are the vertices of a right triangle. The radius of the sphere with center at th
ElenaW [278]

Answer:

r=1

Step-by-step explanation:

First we need to know the length of each side of the triangle, so we use the formula of the vector modulus:

|AB|= \sqrt{(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}+(b_{3}-a_{3})^{2}

By doing so, we find:

|AB|=\sqrt {6}\\|BC|=\sqrt {6}\\|AC|=2\sqrt {5}

With this we know that the triangle is not right, but, we assume the longest side as the hypotenuse of the problem.

As we have two equal sides, we know that the line between point |AB| and the center of the hypotenuse is perpendicular, therefore, we can calculate it using Pythagoras theorem:

|BC|^{2}=r^{2}+(\frac{|AC|}{2})^{2}\\\\r^{2}=|BC|^{2}-(\frac{|AC|}{2})^{2}\\\\r^{2}=(\sqrt{6})^{2}- (\frac{2\sqrt{5}}{2})^{2}\\\\r^{2}=6-5=1\\r=1

4 0
3 years ago
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