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Misha Larkins [42]
3 years ago
11

Darling ohayo!! AAAAAAAAAAAH

Mathematics
2 answers:
Crank3 years ago
3 0

   umm...... ok

free points ig

DanielleElmas [232]3 years ago
3 0

Answer:

freeeee

Step-by-step explanation:

pointss wohooo

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Y = x2 – 3x + 10, find f(–2)
Ann [662]

Answer:Hello

I think it's 12

Step-by-step explanation:

Actually at first you should plus this tow parameters:2x-3x+10

and then you can put(-2)instead (x).

Now you find answer.

6 0
3 years ago
HELP GUYS IM GONNA FAIL MY CLASS HELPPPPPPPPPPPPPP
ollegr [7]

Answer:

SEVEN!!!!

Step-by-step explanation:

5 0
3 years ago
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Please help me I've been stuck on this since the morning. I keep texting my friend to help me but she won't answer me she just l
yan [13]

Answer:

sounds like a relationship problem

Step-by-step explanation:

that cheating wont fix.

4 0
3 years ago
An equation of the perpendicular bisector of the line segment with end points (3,0) and (-3,0) is
Norma-Jean [14]

Answer:

x = 0

Step-by-step explanation:

Given

(x_1,y_1) = (3,0)

(x_2,y_2) = (-3,0)

Required

The equation of the perpendicular bisector.

First, calculate the midpoint of the given endpoints

(x,y) = 0.5(x_1 + x_2, y_1 + y_2)

(x,y) = 0.5(3-3, 0+ 0)

(x,y) = 0.5(0, 0)

Open bracket

(x,y) = (0.5*0, 0.5*0)

(x,y) = (0, 0)

Next, determine the slope of the given endpoints.

m = \frac{y_2 - y_1}{x_2 - x_1}

m = \frac{0 - 0}{-3- 3}

m = \frac{0}{-6}

m = 0

Next, calculate the slope of the perpendicular bisector.

When two lines are perpendicular, the relationship between them is:

m_2 = -\frac{1}{m_1}

In this case:

m = m_1 = 0

So:

m_2 = -\frac{1}{0}

m_2 = unde\ fined

Since the slope is unde\ fined, the equation is:

x = a

Where:

(x,y) = (a,b)

Recall that:

(x,y) = (0, 0)

So:

a = 0

Hence, the equation is:

x = 0

5 0
3 years ago
The graph below shows the function f(x)=x-3/x2-2x-3. Which statement is true?
BARSIC [14]
1. Domain.

We have x^2-2x-3 in the denominator, so:

x^2-2x-3\neq0\\\\(x^2-2x+1)-4\neq0\\\\(x-1)^2-4\neq0\\\\(x-1)^2-2^2\neq0\qquad\qquad[\text{use }a^2-b^2=(a-b)(a+b)]\\\\(x-1-2)(x-1+2)\neq0\\\\
(x-3)(x+1)\neq0\\\\\boxed{x\neq3\qquad\wedge\qquad x\neq-1}

So there is a hole or an asymptote at x = 3 and x = -1 and we know, that answer B) is wrong.

2. Asymptotes:

f(x)=\dfrac{x-3}{x^2-2x-3}=\dfrac{x-3}{(x-3)(x+1)}=\boxed{\dfrac{1}{x+1}}

We have only one asymptote at x = -1 (and hole at x = 3), thus the correct answer is A)
6 0
3 years ago
Read 2 more answers
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