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2 years ago

Solving by elimination -6x+6y=6 -6x+3y=-12

1 answer:

8 0

So elimination method is basically adding the equations and canceling out variables.
-6x + 6y = 6
-6x + 3y = -12
The eaiest way to solve is by multiplying the bottom equation by -1.
-6x + 6y = 6
6x - 3y = 12
Now you add the eqautions.
3y = 18
Divde 3 from both sides.
y = 6
Now plug in 6 into any of the original two equations. Lets use the first one.
-6x + 6(6) = 6
-6x + 36 = 6
Subtract 36 from both sides.
-6x = -30
Divide -6 from both sides.
x = 5
So your solution is (5, 6).
I hope this helps love! :)

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(a) The proportion of accidents that occur up to mile 1.2 of the path is the area under the density curve between 0 miles and 1.

inn [45]

Answer:

the area would be 0.4 or 40%

Step-by-step explanation:

The computation of the area is given below:

But before that the level of the density is

The density level is

= 1 ÷ (3 - 0)

= 1 ÷ 3

Area from 0 miles to 1.2 miles

= (1 ÷ 3) × (1.2 - 0)

= 0.4 or 40 %

Hence, the area would be 0.4 or 40%

the same is considered and relevant too

4 0

3 years ago

Solve for x -2/5x - 8/15x + 1/3x = -54 x=?

Ad libitum [116K]

Answer:

x=0

Step-by-step explanation:

solve for x by simplifying both sides of the equation, then isolating the variable. which got me x=0

3 0

3 years ago

Read 2 more answers

Solve for x. 4−(2x+4)=5 A. x=32 B. x=−52 C. x=−10 D. x = 6

Dennis_Churaev [7]

Answer:

x = - 2.5

Step-by-step explanation:

Hi there !

4 - (2x + 4) = 5

4 - 2x - 4 = 5

- 2x = 5

2x = - 5

x = - 5 : 2

x = - 2.5

Good luck !

6 0

3 years ago

Write an algebraic expression for " the minimum value of 2x + 1 is 13"

bezimeni [28]

2x+1=13 x=6 so an ok answer would be: 2x+10=40 x=15

6 0

3 years ago

FIRST CORRECT ANSWER GETS BRAINLIEST AND 50 POINTS.

AnnyKZ [126]

Answer:

A) $5^3*5^{-4}=5^{3-4}=5^{-1}=\frac{1}{5}$ (Yes. It has a value between zero and one.)

B) $\frac{3^5}{3^{-6}}=3^{5-\left(-6\right)}=3^{5+6}=3^{11}=177147$ (No. It does not have a value between zero and one.)

C) $\left(\frac{1}{4}\right)^3* \left(\frac{1}{4}\right)^2=\left(\frac{1}{4}\right)^{3+2}=\left(\frac{1}{4}\right)^5=\frac{1^5}{4^5}=\frac{1}{1024}$ . (Yes. It has a value between zero and one.)

D) $\frac{\left(-7\right)^5}{\left(-7\right)^7}=\left(-7\right)^{5-7}=\left(-7\right)^{-2}=\frac{1}{\left(-7\right)^2}=\frac{1}{49}$ (Yes. It has a value between zero and one.)

4 0

3 years ago

Read 2 more answers