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3 years ago

PLEASE HELP!!! 15 PTS FOR THIS!! (Please show work also) Convert y=3x-14 (slope intercept form) to standard form

Mathematics

1 answer:

IgorC [24]3 years ago

7 0

Answer:

$3x-y=14$

Step-by-step explanation:

To convert an equation into standard form, we want to get into the form $ax + by = c$ .

If we have $y=3x-14$ , our first step is to get the x term on the left side. We can do this by subtracting 3x from both sides.

$-3x + y = -14$

I find it easier now to just divide both sides by -1, so that we have a more consistent equation.

$3x - y = 14$

Hope this helped!

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What is y= -2/3x + 5 written in standard form?

GaryK [48]

1) the standar form of the equation is ax+by = c, therefore, for y = -2/3x+5, multiply by 3 the previous equation, and its result is 3y = -2x+15⇒ 3y +2x = 15

the final result is 2x+3y = 15

6 0

3 years ago

Read 2 more answers

A dog uses 30 lbs of dog food per day. How many pounds are used in a year?

Ludmilka [50]

To solve this we would do 30×365

We would do this because each day the dog 30lbs

So we have the equation y=30x
Where y= total amount of dog food
And x= number of days

In a year there are 365 days typically, so we do 30×365

30×365=10,950

In one year 10,950 lbs of dog food are used

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3 years ago

A bag of marbles had 14 blue marbles and the rest were white. For a game, 16 marbles were chosen from the bag. Of these 16 marbl

Musya8 [376]

Answer:

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Step-by-step explanation:

6 0

3 years ago

In the illustration below, the three cube-shaped tanks are identical. The spheres in any given tank

fredd [130]

Answer:

1) Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) $Amount \ of \, water \ remaining \ in \, the \ tank \ is \ \frac{x^3(6-\pi) }{6}$

Step-by-step explanation:

1) Here we have;

First tank A

Volume of tank = x³

The volume of the sphere = $\frac{4}{3} \pi r^3$

However, the diameter of the sphere = x therefore;

r = x/2 and the volume of the sphere is thus;

volume of the sphere = $\frac{4}{3} \pi \frac{x^3}{8}$ = $\frac{1}{6} \pi x^3$

For tank B

Volume of tank = x³

The volume of the spheres = $8 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 2·D = x therefore;

r = x/4 and the volume of the sphere is thus;

volume of the spheres = $8 \times \frac{4}{3} \pi (\frac{x}{4})^3= \frac{x^3 \times \pi }{6}$

For tank C

Volume of tank = x³

The volume of the spheres = $64 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 4·D = x therefore;

r = x/8 and the volume of the sphere is thus;

volume of the spheres = $64 \times \frac{4}{3} \pi (\frac{x}{8})^3= \frac{x^3 \times \pi }{6}$

Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) For the 4th tank, we have;

number of spheres on side of the tank, n is given thus;

n³ = 512

∴ n = ∛512 = 8

Hence we have;

Volume of tank = x³

The volume of the spheres = $512 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 8·D = x therefore;

r = x/16 and the volume of the sphere is thus;

volume of the spheres = $512\times \frac{4}{3} \pi (\frac{x}{16})^3= \frac{x^3 \times \pi }{6}$

Amount of water remaining in the tank is given by the following expression;

Amount of water remaining in the tank = Volume of tank - volume of spheres

Amount of water remaining in the tank = $x^3 - \frac{x^3 \times \pi }{6} = \frac{x^3(6-\pi) }{6}$

$Amount \ of \ water \, remaining \, in \, the \ tank = \frac{x^3(6-\pi) }{6}$ .

5 0

3 years ago

F(x) = x^2 - 16 What are the solutions for x when f of x equals 9?

andrey2020 [161]

Answer:

Step-by-step explanation:

5 0

2 years ago