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3 years ago

HELP ASAP Linear Equations

Mathematics

2 answers:

Lubov Fominskaja [6]3 years ago

5 0

Answer:

I and IV only

Step-by-step explanation:

In order for graphs to be parallel, they must have the same slope

In this case, we are looking for a slope of -2/3

We can find the other slopes by solving for y (y=mx+b)

I. 2x+3y=10

Subtract 2x from both sides

3y=-2x+10

Divide by 3

y=-2/3+10/3 This works since the slope is identical

II. 2x-3y=9

Follow a similar progression as the previous (subtract the x's and divide the y's)

-3y=-2x+9

y=2/3=3 This doesn't work since it is positive

III. 4x-6y=30

-6y=-4x+30

y=2/3 x-5 (4/6 simplified to 2/3) Also doesn't work since it is positive

IV. 4x+6y=3

6y=-4x+3

y=-2/3x+3 (-4/6 simplified to -2/3) This one works, it is identical

Scilla [17]3 years ago

4 0

Answer:

4th choice: I and IV only

Step-by-step explanation:

Change the form of the given equation to standard form.

y = -2/3 x + 2

3y = -2x + 6

2x + 3y = 6

Now compare this equation with the 4 choices.

Any equation that has the same x- and y-terms but a different constant term is parallel.

I. is parallel.

II. and III. are not parallel.

Look at IV.:

4x + 6y = 3

Divide both sides by 2:

2x + 3y = 3/2

IV. is also parallel.

Answer: 4th choice: I and IV only

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Anarel [89]

Given:

The figure of a right angle triangle.

$a=b$

Hypotenuse = $9\sqrt{2}$ in.

To find:

The missing lengths of the sides.

Solution:

In the given right angle triangle both legs a and b are equal, and hypotenuse is $9\sqrt{2}$ in.

Using Pythagoras theorem, we get

$Hyponteuse^2=Base^2+Perpendicular^2$

$(9\sqrt{2})^2=(a)^2+(b)^2$

$81(2)=(a)^2+(a)^2$ $[\because a=b]$

$81(2)=2(a)^2$

Divide both sides by 2.

$81=(a)^2$

Taking square root on both sides.

$\pm \sqrt{81}=a$

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Side cannot be negative. So,

$9=a$

Thus, the missing side lengths are a=9 in and b=9 in.

Therefore, the correct option is C.

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