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3 years ago

An angle measures 51 , what is the measure of its complement

Mathematics

1 answer:

n200080 [17]3 years ago

8 0

Answer:

complement of 51 is 39

supplement of 54 is 126

Step-by-step explanation:

Complementary angles add to 90 degrees

51 + c = 90

Subtract 51 from each side

51-51 + c = 90-51

c = 39

The complement is 31

Supplementary angles add to 180 degrees

54 + s = 180

Subtract 54 from each side

54-54 + s = 180-54

s = 126

The supplement is 126

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Use angle relationships to complete the top equation. Then, solve for x and find the measure of both angles.

Ad libitum [116K]

Answer:

Step-by-step explanation:

These two angles are interior and on the same side of the transversal. Therefore they are supplementary.

6x + x + 5 = 180

7x + 5 = 180 Subtract 5 from both sides

7x = 180 - 5

7x = 175 Divide by 7

x = 175/7

x = 25

x + 5 = 25 + 5 = 30

6x = 6*25 = 150

6 0

3 years ago

Identify the expression that is not equal to the other three. Explain your reasoning

Nadusha1986 [10]

Answer:

$-\left|7+3\right|$

Step-by-step explanation:

Solve each expression and then compare the results

case 1) we have

$15-\left|-5\right|\\ \\=15-(5)\\ \\=10$

case 2) we have

$\left|-4\right|+6\\ \\=(4)+6\\ \\=10$

case 3) we have

$-\left|7+3\right|\\ \\ =-(7+3)\\ \\=-10$

case 4) we have

$\left|-10\right|\\ \\ =-(10)\\ \\=10$

3 0

3 years ago

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ValentinkaMS [17]

Answer:

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Step-by-step explanation:

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8 0

3 years ago

Three odd-shaped blocks of chocolate have the following masses and center-of-mass coordinates: (1) 0.310kg , ( 0.200 m, 0.320m )

Katarina [22]

Answer:

The x coordinate is 0.047m while y coordinate is 0.081 m

Step-by-step explanation:

The coordinates of the center of the mass of the system are given as follows

$x_c=\dfrac{\sum_{i=1}x_im_i}{\sum_{i=1}m_i}\\y_c=\dfrac{\sum_{i=1}y_im_i}{\sum_{i=1}m_i}$

So from the data

m1=0.310, x1=0.20,y1=0.320

m2=0.410,x2=0.110,y2=-0.380

m3=0.210,x3=-0.300,y3=0.630

Now by substitution of the values in the above equation

$x_c=\dfrac{\sum_{i=1}x_im_i}{\sum_{i=1}m_i}\\x_c=\dfrac{x_1m_1+x_2m_2+x_3m_3}{m_1+m_2+m_3}\\x_c=\dfrac{0.310*0.2+0.410*.110+0.210*-0.3}{0.310+0.410+0.210}\\x_c=0.047m\\\\\\y_c=\dfrac{\sum_{i=1}y_im_i}{\sum_{i=1}m_i}\\y_c=\dfrac{y_1m_1+y_2m_2+y_3m_3}{m_1+m_2+m_3}\\y_c=\dfrac{0.310*0.320+0.410*-0.380+0.210*.630}{0.310+0.410+0.210}\\y_c=0.081m$