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Sergeu [11.5K]
2 years ago
5

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

Mathematics
1 answer:
Brums [2.3K]2 years ago
5 0

Answer:

-1/4

Step-by-step explanation:

y1 y2    x1 x2

17 - 18=-1      11 - (-15 ) 4

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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
11 months 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
2 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
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