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3 years ago

Find the rate of change for this linear function. y= - 8x+1

Mathematics

2 answers:

OLga [1]3 years ago

8 0

Answer:

-8

slope is -8/1

m= -8

Step-by-step explanation:

kakasveta [241]3 years ago

8 0

Yes it was crazy due to the fact that I’m doing this just to complete step 4 of my stuff

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Find p(-4) and p(2) for the function p(x)=11x^5-11x^4-5x^2+15x-8

egoroff_w [7]

hi,

you must replace x by the number between parenthese.

I show you with the first one and let you do the second one

p(x) = 11x^5 -11x^4 - 5x^2 +15x-8

p(-4) = 11 (-4)^5 - 11 (-4)^4 -5(-4)² +15(-4) -8

p(-4) = 11 ( -1024- 256) - 5*16 -60-8

p(-4) = 11 ( -1280) -80-60-8

p(-4) = - 14080 - 148

p(-4) = - 14 228

7 0

2 years ago

Hey! pls help i’ll give brainliest

grandymaker [24]

Answer:

landslide

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

12) Erica has three more dimes than nickels in her pocket, for a total of $1.50. If x represents the number of

serious [3.7K]

Answer:

Step-by-step explanation:

lol

8 0

3 years ago

Read 2 more answers

Write an equivalent expression for 7 + 5 k minus 2 minus 3 k + n. Which statements are true about the steps for writing the equi

Nataly [62]

Answer:

1. Combine the constant terms by adding 7 and –2.

4. Combine the like variable terms by adding the coefficients

5. 5 k and negative 3 k are like variable terms

6.The constants 7 and –2 are like terms.

7. The equivalent expression is 5 + 2 k + n.

5 0

2 years ago

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

3 0

2 years ago