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3 years ago

In the scalene triangles shown, 2P = 2).

Mathematics

1 answer:

Black_prince [1.1K]3 years ago

8 0

Answer:

I guess (c) is your answer. thanks!!

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Which function has the graph shown? <br> PLEASE HELP

elena-14-01-66 [18.8K]

Try graphing them in desmos

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3 years ago

A line passes through (−1, 7) and (2, 10). which answer is the equation of the line?

mart [117]

First you need to find the slope of the line. So what you need to do is y^2-y^1 / x^2-x^1

(x^1, y^1) (x^2, y^2)
( -1, 7) ( 2, 10)

10-7/ 2- (-1)
3/3=1
slope is 1

I got 8 as the y intercept. Reason how is that I made a table. And what I did was go back instead of going forward.

x y
-1 7
0 8
1 9
2 10

The answer would be x= 1 y= x + 8 y=8 But the answer choice you listed for A. would not match up the 2 coordinate you gave. Make sure a. should be x= 1 y= x + 8 y=8, not -x y= 8-x y=8

6 0

3 years ago

72.6 meters in 11 seconds

ira [324]

This would simplify to 6.60 meters/second.

8 0

3 years ago

Read 2 more answers

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

3 years ago

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Use the drawing tools to form the correct answers on the graph. Plot the vertex and the axis of symmetry of this function: f(x)

CaHeK987 [17]

Answer:

Axis of Symmetry: x = 3

Vertex: (3, 5)

Step-by-step explanation:

Use a graphing calc.

6 0

3 years ago

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