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2 years ago

Please help me with complement or supplement of an angles.

Mathematics

1 answer:

rosijanka [135]2 years ago

6 0

Answer:

Step-by-step explanation:

If we look at the picture, we can see that its a right angle having a 90 degree.

We can make a equation with this as we know that m1 and m2 has to equal 90

x + y = 90

Since we know that m2 = 30 we can plug 30 for x

(30) + y = 90

Now we simply moving the 30 to the 90

y = 90 - 30

y = 60

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59-(2c+3)=4(c+7)+c.

alisha [4.7K]

Step-by-step explanation:

${:}\leadsto$ $\sf 59-(2c+3)=4 (c+7)+c$

${:}\leadsto$ $\sf 59-2c-3=4c+28+c$

${:}\leadsto$ $\sf 56-2c=5c+28$

${:}\leadsto$ $\sf 5c+28=56-2c$

${:}\leadsto$ $\sf 5c+2c=56-28$

${:}\leadsto$ $\sf 7c=28$

${:}\leadsto$ $\sf c={\dfrac {28}{7}}$

${:}\leadsto$ $\sf c=4$

4 0

2 years ago

Read 2 more answers

Evaluate the expression using the values in the left column to complete the table

egoroff_w [7]

Answer/Step-by-step explanation:

A. $4b + 3b$

Using b = 2

$4(2) + 3(2) = 8 + 6 = 14$

Using b = 4

$4(4) + 3(4) = 16 + 12 = 28$

Using b = 6

$4(6) + 3(6) = 24 + 18 = 42$

Using b = 8

$4(8) + 3(8) = 32 + 24 = 56$

Using b = 10

$4(10) + 3(10) = 40 + 30 = 70$

Using b = 2

$4(12) + 3(12) = 48 + 36 = 84$

B. $3(x - 2)$

Using x = 3

$3(3 - 2) = 3(1) = 3$

Using x = 5

$3(5 - 2) = 3(3) = 9$

Using x = 7

$3(7 - 2) = 3(5) = 15$

Using x = 9

$3(9 - 2) = 3(7) = 21$

Using x = 11

$3(11 - 2) = 3(9) = 27$

Using x = 13

$3(13 - 2) = 3(11) = 33$

C. $5(4a - 1)$

Using a = 2

$5(4(2) - 1) = 5(8 - 1) = 5(7) = 35$

Using a = 3

$5(4(3) - 1) = 5(12 - 1) = 5(11) = 55$

Using a = 4

$5(4(4) - 1) = 5(16 - 1) = 5(15) = 75$

Using a = 5

$5(4(5) - 1) = 5(20 - 1) = 5(19) = 95$

Using a = 6

$5(4(6) - 1) = 5(24 - 1) = 5(23) = 115$

Using a = 7

$5(4(7) - 1) = 5(28 - 1) = 5(27) = 135$

D. $6y - 2y + 4$

Using y = 2

$6(2) - 2(2) + 4 = 12 - 4 + 4 = 12$

Using y = 3

$6(3) - 2(3) + 4 = 18 - 6 + 4 = 18 - 2 = 16$

Using y = 5

$6(5) - 2(5) + 4 = 30 - 10 + 4 = 30 - 6 = 24$

Using y = 10

$6(10) - 2(10) + 4 = 60 - 20 + 4 = 60 - 16 = 44$

Using y = 8

$6(8) - 2(8) + 4 = 48 - 16 + 4 = 48 - 12 = 36$

Using y = 6

$6(6) - 2(6) + 4 = 36 - 12 + 4 = 36 - 8 = 28$

7 0

3 years ago

What is the answer to this too?

Radda [10]

The answer to this is 5%

4 0

2 years ago

How to prove two equations have infinite solutions?

laiz [17]

To prove two equations have infinite solutions, you have to prove that those two equations are the same equations, but in a different form.

For example: Prove the equations are infinite
5y=2x+7
10y=4x+14
If you multiply the first equation by 2, and substitiute any of the numbers, you will get 0=0

6 0

2 years ago

Find all critical values of the function g(x)=5/(x²-x+6)

Sati [7]

$g(x)=\frac{5}{x^2-x+6}\\\\D:x^2-x+6\neq0\\\\\Delta=(-1)^2-4\cdot1\cdot6=1-24=-23 < 0\\\\x\in\mathbb{R}\\\\vertex\ of\ y=x^2-x+6\\\\p=\frac{-(-1)}{2\cdot1}=\frac{1}{2}\\\\q=\frac{-(-23)}{4\cdot1}=\frac{23}{4}\\\\\lim\limits_{x\to\frac{1}{2}}\frac{5}{x^2-x+6}=\frac{5}{\frac{23}{4}}=\frac{5\cdot4}{23}=\frac{20}{23}$

$\lim\limits_{x\to\pm\infty}\frac{5}{x^2-x+6}=0$

3 0

3 years ago