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2 years ago

Find the slope of the line that passes through each pair of points

Mathematics

1 answer:

lyudmila [28]2 years ago

3 0

Answer:

-1

Step-by-step explanation:

The slope of the line can be found by

m = (y2-y1)/(x2-x1)

= (-2-1)/(1--2)

=-3 /(1+2)

=-3/3

-1

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andreyandreev [35.5K]

Answer:

Step-by-step explanation:

$\dfrac{\frac{2}{3}}{\frac{1}{3}}=\frac{2}{3}\cdot \frac{3}{1}=\frac{6}{3}=2$ . Hope this helps!

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3 years ago

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PLs help 50 PTS!!!!! PLEASE ILL GIVE BRAINLIEST!!!!!

Nookie1986 [14]

Answer:

$\large\boxed{y=\dfrac{1}{4}x^2-x-4}$

Step-by-step explanation:

The equation of a parabola in vertex form:

$y=a(x-h)^2+k$

(h, k) - vertex

The focus is

$\left(h,\ k+\dfrac{1}{4a}\right)$

We have the vertex (2, -5) and the focus (2, -4).

Calculate the value of a using $k+\dfrac{1}{4a}$

k = -5

$-5+\dfrac{1}{4a}=-4$ add 5 to both sides

$\dfrac{1}{4a}=1$ multiply both sides by 4

$4\!\!\!\!\diagup^1\cdot\dfrac{1}{4\!\!\!\!\diagup_1a}=4$

$\dfrac{1}{a}=4\to a=\dfrac{1}{4}$

Substitute

$a=\dfrac{1}{4},\ h=2,\ k=-5$

to the vertex form of an equation of a parabola:

$y=\dfrac{1}{4}(x-2)^2-5$

The standard form:

$y=ax^2+bx+c$

Convert using

$(a-b)^2=a^2-2ab+b^2$

$y=\dfrac{1}{4}(x^2-2(x)(2)+2^2)-5\\\\y=\dfrac{1}{4}(x^2-4x+4)-5$

use the distributive property: a(b+c)=ab+ac

$y=\left(\dfrac{1}{4}\right)(x^2)+\left(\dfrac{1}{4}\right)(-4x)+\left(\dfrac{1}{4}\right)(4)-5\\\\y=\dfrac{1}{4}x^2-x+1-5\\\\y=\dfrac{1}{4}x^2-x-4$

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3 years ago

1000x8129= WILL MAKR BRANLYEST

8_murik_8 [283]

Answer:

it is 8,129,000

Step-by-step explanation:

1000x8129 = 8,129,000

8 0

2 years ago

Please Help!!!!!! it's due this morning

zhuklara [117]

Answer:

Check the solution in both equations. The solution is ( − 1, 2). Solve the system by graphing: {− x + y = 12 x + y = 10 . Solve the system by graphing: {2x + y = 6 x + y = 1 . In all the systems of linear equations so far, the lines intersected and the solution was one point.

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2 years ago

Help please thank you!

monitta

Answer:

• multiplied by 4p: (x -h)² +4pk = 0

• zeros for k > 0: none

• zeros for k = 0: one

• zeros for k < 0: two

Step-by-step explanation:

a) Multiplying by 4p removes the 1/(4p) factor from the squared term, but adds a factor of 4p to the k term. (It has no effect on the subsequent questions or answers, so we wonder why we're doing this.) The result is ...

(x -h)² +4pk = 0

b) The value of k is the vertical location of the vertex of the parabola with respect to the x-axis. The parabola opens upward, so for k > 0, the parabola does not cross the x-axis, and the number of real zeros is zero. (There are two complex zeros.)

c) As in part b, the value of k defines the vertex location. When it is zero, the vertex of the parabola is on the x-axis, so there is one real zero (It is considered to have multiplicity 2.)

d) As in part b, the value of k defines the vertex location. When it is negative, the vertex of the parabola is below the x-axis. Since the parabola opens upward, both branches will cross the x-axis, resulting in two real zeros.

_____

The attached graph shows a parabola with p=1/4 and h=2. The values shown for k are +1, 0, and -1. The coordinates of the real zeros are shown.

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3 years ago