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3 years ago

8

Hellpppp I’m stuck......

2 answers:

Dafna1 [17]3 years ago

3 0

I’m just gonna guess but I think 6

Anettt [7]3 years ago

3 0

i think it’s five
Vertical angles are angles that congruent and opposite of each other

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Solve the system by elimination.(show your work)

PilotLPTM [1.2K]

Answer:

x = 1 , y = 1 , z = 0

Step-by-step explanation by elimination:

Solve the following system:

{-2 x + 2 y + 3 z = 0 | (equation 1)

-2 x - y + z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x - 3 y - 2 z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y+0 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1 , y = 1 , z = 0

6 0

3 years ago

Read 2 more answers

Pls send help my waayyy

soldier1979 [14.2K]

Answer:

B

Step-by-step explanation:

110 / 22 = 5 inches per picket

equation is y = 5x

3 0

4 years ago

Oshua's mother has a family Christmas gathering at 2:00. Joshua must travel 348 miles to join the festivities. He left his house

gavmur [86]

Answer:

58 miles/hour

Step-by-step explanation:

Given that the family Christmas gathering has been scheduled at 2:00 p.m.,

and Joshua left the house at 8:00 a.m.

So, the total time available in order to arrive on time to attend the Christmas gathering, t=6 hours

The total distance, Joshua has to travel, d= 348 miles.

As average speed= (total distance)/(total time)

So, the average speed of driving = 348/6=58 miles/hour

Hence, Joshua must drive at an average speed of 58 miles/hour in order to arrive on time.

7 0

3 years ago

Use the distributive property to write an expression that is equivalent to each given property.

Nadya [2.5K]

Answer:

-6x+12

-9+5a

7x+63

8y-4/5x-12

Step-by-step explanation:

To solve these all you have to do is take the number outside of the parenthesis and multiply them by each term.

-3(2x)-3(-4) All I did here was expand the problem to show what i mean by terms.

-6x+12

Do this for each problem.

0.1(-90)+0.1(50a)

-9+5a

-7(-x)-7(-9) For this one you are taking negatives times negatives so each answer will be positive.

7x+63

4/5(10y)+4/5(-x)+4/5(-15)

8y-4/5x-12

3 0

3 years ago

Solve the following equation by factoring:9x^2-3x-2=0

olya-2409 [2.1K]

Answer:

The two roots of the quadratic equation are

$x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}$

Step-by-step explanation:

Original quadratic equation is $9x^{2}-3x-2=0$

Divide both sides by 9:

$x^{2} - \frac{x}{3} - \frac{2}{9}=0$

Add $\frac{2}{9}$ to both sides to get rid of the constant on the LHS

$x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}$ ==> $x^{2} - \frac{x}{3}=\frac{2}{9}$

Add $\frac{1}{36}$ to both sides

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}$

This simplifies to

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}$

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = $\frac{1}{6}\right)$ we can see that

$\left(x - \frac{1}{6}\right)^2$ = $x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}$

So

$\left(x - \frac{1}{6}\right)^2=\frac{1}{4}$

Taking square roots on both sides

$\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}$

So the two roots or solutions of the equation are

$x - \frac{1}{6}=-\sqrt{\frac{1}{4}}$ and $x - \frac{1}{6}=\sqrt{\frac{1}{4}}$

$\sqrt{\frac{1}{4}} = \frac{1}{2}$

So the two roots are

$x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}$

and

$x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}$

7 0

2 years ago