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3 years ago

Write an equation in slope-intercept form for the line with slope 7 and y-intercept -7 .

1 answer:

3 0

Y=7x-7

The slope intercept form would be y=Mx+b

The m in the equation is the slope
The b in the equation is the y-intercept

You can substitute these values in the intercept form.

y=Mx+b
y=(7)x+(-7)
y=7x-7

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How do i do this? i'm so lost. please explain and show your work

elena-s [515]

You want to make sure to factor each equation as much as you can first.
So just focus on the numerator first. $x^3-6x^2+8x$
To factor the numerator:
$(x)(x^2-6x+8)$ factor out an x
$(x)(x-4)(x-2)$ factor out the rest completely.

Now for the denominator: $-4x^2+4x+24$
$(-4)(x^2+x+6)$ factor out a -4
$(-4)(x+3)(x-2)$ factor out the rest.

This leaves you with:
$\frac{(x)(x-4)(x-2)}{(-4)(x+3)(x-2)}$

Then you can simplify:
$\frac{(x)(x-4)}{(-4)(x+3)}$

I'm not sure what you're supposed to find exactly, so if you need more help you can ask me (:

8 0

4 years ago

Please help my sister with this! 4. 3 gallons of paint are used to paint 16 wooden signs how much paint did you sign used betwee

Paul [167]

Answer:

Step-by-step explanation:

It would be 16 divided by 3 and from the answer you get from that you should divide it by itself equaling infinite .111111111111etc

4 0

3 years ago

If a ball is thrown into the air with a velocity of 34 ft/s, its height (in feet) after t seconds is given by y = 34t − 16t2. Fi

Finger [1]

<u>ANSWER: </u>

If a ball is thrown into the air with a velocity of 34 feet per second, then velocity of the ball after 1 second is 2 feet per second

<u>SOLUTION: </u>

Given, a ball is thrown into the air with a velocity of 34 feet per second

Initial velocity (u) = 34 feet per second

And also given a relation between displacement and time = $\mathrm{y}=34 \mathrm{t}-16 \mathrm{t}^{2}$ --- eqn 1

We need to find the velocity when t = 1 ; v = ?

We know that, v = u + at and $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

where v is instantaneous velocity and u is initial velocity

a is acceleration

t is time interval

s is displacement

using the displacement and time relation eqn (1) we get

Now, when t = 1, displacement s = 34(1) – 16(1)

$\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}=34-16$

$34 \times 1+\frac{1}{2} \times a \times 1^{2}=18$

$34+\frac{a}{2}=18$

$\begin{array}{l}{\frac{a}{2}=18-34} \\\\ {\frac{a}{2}=-16} \\ {a=-16 \times 2} \\ {a=-32}\end{array}$

here, -ve sign indicates that object is in deceleration . so acceleration is -32 ft/s

now put a value in v = u + at

v = 34 + (-32)(1)

v = 34 – 32

v = 2 ft/s

Hence, velocity of the ball after 1 second is 2 ft/s

6 0

3 years ago

Sam is eating a Big Hamburger. The first bite was 20% of the Hamburger, the second bite was 20% of what is left and so every nex

lakkis [162]

The weight of the hamburger is 160g after two bites.

Each bite takes away 20% of the weight of the burger, so we'll divide the initial weight by a percentage to find the weight before the bite.

The starting weight is 160g. Because 20% of the weight was removed from a bite, we'll divide this weight by the percentage of the burger remaining after the bite:

$100 - 20 = 80$

80% of the burger is left after a bite. Convert this fraction into a decimal by dividing by 100:

$80 \div 100 = 0.8$

Divide the initial weight by this decimal:

$160 \div 0.8 = 200$

The weight of the burger before the second bite was 200.

We'll do the same procedure for the first bite. The initial weight is 200, and there is 80% of the burger remaining after the first bite. Because 80% is 0.8 in decimal form, divide 200 by 0.8:

$200 \div 0.8 = 250$

The starting weight of the burger before the bites was 250 grams.

6 0

3 years ago

Are the ratios 3:6 and 6:3 equivalent?

KATRIN_1 [288]

Yes they are cause 3 of 6 and 6 of 3

4 0

3 years ago

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