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Tomtit [17]
2 years ago
11

Theo's friend made a display with dimensions that were each half as long as those theo used. what is the approximate mass of the

o's friends display
Mathematics
1 answer:
Crank2 years ago
4 0

Answer:

great question buddy and I wish you the most

Step-by-step explanation:

x-y=b^2/2=2

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2. One angle is one less than six times the measure of
Illusion [34]

The equations used to find the measure of each angle in degrees is x + y = 90 and x = 6y - 1

The two complementary angles are 77 degrees and 13 degrees

<em><u>Solution:</u></em>

Given that two angles are complementary angles

Complementary angles are two angles whose sum is 90 degrees

Let one of the angle be "x" and the other angle be "y"

Therefore,

x + y = 90 ------ eqn 1

Also given that,

One angle is one less than six times the measure of  another

one angle = six times the other angle - 1

x = 6y - 1 ------ eqn 2

Substitute eqn 2 in eqn 1

6y - 1 + y = 90

Thus the above equation is used to find the measure of each angle in degrees

Solve the above equation

6y + y - 1 = 90

7y - 1 = 90

7y = 91

y = 13

Substitute y = 13 in eqn 2

x = 6(13) - 1

x = 78 - 1

x = 77

Thus the two complementary angles are 77 degrees and 13 degrees

8 0
3 years ago
An object is launched from ground level with an initial velocity of 120 meters per second. For how long is the object at or abov
babunello [35]

Answer:

D) 14 seconds

Step-by-step explanation:

First we will plug 500 in for y:

500 = -4.9t² + 120t

We want to set this equal to 0 in order to solve it; to do this, subtract 500 from each side:

500-500 = -4.9t² + 120t - 500

0 = -4.9t²+120t-500

Our values for a, b and c are:

a = -4.9; b = 120; c = -500

We will use the quadratic formula to solve this.  This will give us the two times that the object is at exactly 500 meters.  The difference between these two times will tell us when the object is at or above 500 meters.

The quadratic formula is:

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Using our values for a, b and c,

x=\frac{-120\pm \sqrt{120^2-4(-4.9)(-500)}}{2(-4.9)}\\\\=\frac{-120\pm \sqrt{14400-9800}}{-9.8}\\\\=\frac{-120\pm \sqrt{4600}}{-9.8}\\\\=\frac{-120\pm 67.8233}{-9.8}\\\\=\frac{-120+67.8233}{-9.8}\text{ or }\frac{-120-67.8233}{-9.8}\\\\=\frac{-57.1767}{-9.8}\text{ or }\frac{-187.8233}{-9.8}\\\\=5.3242\text{ or }19.1656

The two times the object is at exactly 500 meters above the ground are at 5 seconds and 19 seconds.  This means the amount of time it is at or above 500 meters is

19-5 = 14 seconds.

6 0
3 years ago
What are two ratios that are equivalent to 27:9
skad [1K]
For this case we have the following relationship:
 27:9
 To find two equivalent relationships, what we must do is divide both numbers by a number that is multiple of both.
 We then have to divide by three:
 9:3
 Then, dividing again between three we have:
 3:1
 Answer:
 two ratios that are equivalent to 27:9 are:
 Ratio 1:
 9:3
 Ratio 2:
 3:1
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Has anyone taken Ap language and composition
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Answer:no

Step-by-step explanation:

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