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alexandr402 [8]

3 years ago

11

Are the angles with the given measures complementary, supplementary, or neither?

1 answer:

Vlada [557]3 years ago

4 0

The two angles are supplementary angles. Two angles are considered to be supplementary when the sum of the measurement of their angles is 180 degrees. The measurement of angle T is 76 degrees and the measurement of angle U is 104 degrees. If you try to add the measurement of the two angles, the answer would be 180 degrees. Therefore, the two angles are supplementary angles.

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Which algebraic expression is equivalent to the expression below 6(2x+5)+4x

Veronika [31]

12x+30+4
12x+34

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Free coins- 1) what is 2+2?

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Step-by-step explanation:

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mafiozo [28]

Answer:

Step-by-step explanation:

a. 16b^2c^12-0.25 can be rewritten as follows because it is the difference of two squares:

(4bc^6 - .5)(4bc^6 + .5)

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is the difference of two squares, just like (a) (above); its factors are:

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7 0

3 years ago

Classify by degree and number of terms, in that order<br><br> 5x^2-2x+3

zmey [24]

Answer:

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6 0

3 years ago

A 4-lb. force acting in the direction of (vector) 4,-2 moves an object just over 7ft from point (0,4) to (5,-1). Find the work d

Tcecarenko [31]

To solve this problem, we have to find the net displacement and the net force and the multiply the dot product together and get the work done.

The work done on moving the object is 27ft*lbs

<h3>Work done in moving the object from point A to point B</h3>

To find the work done on this object, let's find the net force on the object.

Data;

force = 4lb
direction = 4, -2
displacement = 7ft
direction = (0, 4) to (5,1)

The unit vector of the force is

$\sqrt{4^2 +(-2)^2} =\sqrt{16 + 4} = \sqrt{20}$

$\frac{4}{\sqrt{20} }, \frac{-2}{\sqrt{20} }$

The net force acting on the object is

$F = 4(\frac{4}{\sqrt{20} }, \frac{-2}{\sqrt{20} })\\F= (\frac{16}{\sqrt{20} }, \frac{-8}{\sqrt{20} } )$

The displacement on the object is 7ft through (0,4) to (5, -1)

The unit vector on displacement is

$\sqrt{5^2 + (-1-4)^2} = \sqrt{25+25} = \sqrt{50}$

$\frac{5}{\sqrt{50} }, \frac{-5}{\sqrt{50} }$

The net displacement will be

$7(\frac{5}{\sqrt{50} }, \frac{-5}{\sqrt{50} }) = \frac{35}{\sqrt{50} }, \frac{-35}{50}$

The work done will be F.d

$w = f. d \\$

$w = (\frac{16}{\sqrt{20} }, \frac{-8}{\sqrt{20} } ) * \frac{35}{\sqrt{50} }, \frac{-35}{50}\\w = 17.71+ 8.854\\w = 26.567 = 27ft*lbs$

The work done on moving the object is 27ft*lbs

Learn more on work done on an object here;

brainly.com/question/26152883

#SPJ1

6 0

2 years ago