Ask question

Ask question

All categories

3 years ago

5

Can you help me find the point slope form (-15,18) (11,17)

1 answer:

Brums [2.3K]3 years ago

5 0

Answer:

-1/4

Step-by-step explanation:

y1 y2 x1 x2

17 - 18=-1 11 - (-15 ) 4

You might be interested in

X^2+y^2=1 yes or no relation represents a function or not

sleet_krkn [62]

We are asked to decide if the expression:

x^2 + y^2 = 1 represents a function.

We recall that in order to have a function, we need for a given value of x to have a SINGLE value of y associated with it.

So in this case, when x is 0 for example, we have the following:

0^2 + y^2 = 1

then y^2 = 1

and we realize that there are TWO values of y whose square form gives 1 (one is 1 and the other -1) Therefore, this relationship is NOT a function, since for example when x = 0 there are TWO values of y to which that x is associated (y = 1 and y = -1).

So please select that this is NOT a function for your answer.

6 0

1 year ago

In the next 10 months, Colin wants to save $900 for his vacation. He plans to save $75 each of the first 8 months. How much must

Ivanshal [37]

He needs to save $150 each of the last two months.

5 0

3 years ago

Please help me thanks

PilotLPTM [1.2K]

6x + 45 + 3x = 180
9x - 45 = 180
9x = 135
x = 15

8 0

2 years ago

Mrs. Metheney had some donuts. She bought 8 more, and now has 18 in total. Write the equation using d as your variable

Tpy6a [65]

Answer:

y=10d+8

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Find the product of all real values of r for which 1/2x=r-x/7

Dahasolnce [82]

Answer:

$r = \±\sqrt{14$

$Product = -14$

Step-by-step explanation:

Given

$\frac{1}{2x} = \frac{r - x}{7}$

Required

Find all product of real values that satisfy the equation

$\frac{1}{2x} = \frac{r - x}{7}$

Cross multiply:

$2x(r - x) = 7 * 1$

$2xr - 2x^2 = 7$

Subtract 7 from both sides

$2xr - 2x^2 -7= 7 -7$

$2xr - 2x^2 -7= 0$

Reorder

$- 2x^2+ 2xr -7= 0$

Multiply through by -1

$2x^2 - 2xr +7= 0$

The above represents a quadratic equation and as such could take either of the following conditions.

(1) No real roots:

This possibility does not apply in this case as such, would not be considered.

(2) One real root

This is true if

$b^2 - 4ac = 0$

For a quadratic equation

$ax^2 + bx + c = 0$

By comparison with $2x^2 - 2xr +7= 0$

$a = 2$

$b = -2r$

$c =7$

Substitute these values in $b^2 - 4ac = 0$

$(-2r)^2 - 4 * 2 * 7 = 0$

$4r^2 - 56 = 0$

Add 56 to both sides

$4r^2 - 56 + 56= 0 + 56$

$4r^2 = 56$

Divide through by 4

$r^2 = 14$

Take square roots

$\sqrt{r^2} = \±\sqrt{14$

$r = \±\sqrt{14$

Hence, the possible values of r are:

$\sqrt{14$ or $-\sqrt{14$

and the product is:

$Product = \sqrt{14} * -\sqrt{14}$

$Product = -14$

8 0

3 years ago