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4 years ago

What is the equation in point slope form for the given points and slope .1 and 9 slope 5

Mathematics

1 answer:

Musya8 [376]4 years ago

7 0

Y=mx+b (1,9) and slope= 5 9=5(1)+b 9=5+b 9-5=b 4=b Equation is Y=5x+4

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A recipe calls for 4 sliced apples. You cut the apples into thirds. How many pieces of apple do you have? A) 3 pieces B) 6 piece

OverLord2011 [107]

When you cut the apples into thirds, each apple gives you 3 pieces of equal size.

1 apple = 3 pieces

4 * 1 apple = 4 * 3 pieces

4 apples = 12 pieces

4 0

4 years ago

Read 3 more answers

Which figure accurately represents the Pythagorean theorem?

Elena-2011 [213]

Figure B is the answer!

The formula for the Pythagorean theorem is

A^2+B^2=C^2

In figure B the units are squared. Making it your answer.

-Seth

8 0

4 years ago

What is the slope of the line that passes through the points (2, 1)(2,1) and (17, -17) ?(17,−17)? Write your answer in simplest

Hunter-Best [27]

$\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{17}~,~\stackrel{y_2}{-17}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-17}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{17}-\underset{x_1}{2}}}\implies \cfrac{-18}{15}\implies -\cfrac{6}{5}$

8 0

3 years ago

If f(x) = 2x2 + 3 and g(x)=x2-7, find (f- g)(x).

Len [333]

The difference of the given functions is x^2 + 10

<h3>Difference of functions</h3>

Given the following functions

f(x) = 2x^2 + 3 and

g(x)=x^2-7

Take the difference

(f-g)(x) = 2x^2 + 3 - (x^2 - 7)

(f-g)(x) = 2x^2 - x^2 +3 + 7

(f-g)(x) = x^2 + 10

Hence the difference of the given functions is x^2 + 10

Learn more on difference of function here:brainly.com/question/17431959

#SPJ1

5 0

2 years ago

Use the limit theorem and the properties of limits to find the horizontal asymptotes of the graph of the function g(x) = x2/x2-2

ElenaW [278]

Answer:

Option b is right.

Step-by-step explanation:

A function is given as

$g(x) = \frac{x^2}{x^2-2x-1}$

Limit is to be found out for x tends to infinity.

We find that numerator and denominator has the same degree.

HEnce a horizontal asymptote exists

COefficients of leading terms are 1 and 1 respectively

Asymtote would be y =1/11 = 1

Alternate method:

When x tends to infinity, 1/x tends to 0

$g(x) =\frac{\frac{1}{x^2} }{1-\frac{2}{x} -\frac{1}{x^2} }$

by dividing both numerator and denominator by square of x.

Now take limit as 1/x tends to 0

we get

limit is y tends to 1/1 =1

Hence horizontal asymptote is y =1

8 0

4 years ago