Ask question

Ask question

All categories

3 years ago

11

write the equation of the line that passes through the point (-2,4) and (5,8). Put your answer in fully reduced point slope form

, unless it is a vertical or horizontal line.

1 answer:

sergij07 [2.7K]3 years ago

8 0

Answer:

y-4=4(x+2)/7

Step-by-step explanation:

slope(m) = (8-4)/(5-(-2)) =4/7

b = 4-(4/7)×(-2) = 36/7

y = mx+b

y = 4x/7+36/7

in point-slope form,

y-4=4(x+2)/7

You might be interested in

Please help. I don't understand this. Thanks.

neonofarm [45]

Yes;k=-3 and y=-3x is the answer because 2 x -3 = -6 -3 x -3 = -9 and etc..

4 0

3 years ago

peaches are being sold go $2 per pound. The customer created a model to represent the total cost of peaches bought. if x represe

kvv77 [185]

Answer:

y=2x

Step-by-step explanation:

x independent?

y dependent

positive change? so srry

4 0

2 years ago

Two angles are supplementary. The larger angle is 33 degrees more than 6 times the smaller angle. Find the measure of each angle

noname [10]

Sum of two angles that are supplementary = 180°

Let the smaller angle be = x

$Let \: the \: smaller \: angle \: be \: = \: x$

$then \: the \: larger \: angle \: = 6x + 33$

<h3>Their sum :</h3>

$x + 6x + 33 = 180$

$7x + 33 = 180$

$7x = 180 - 33$

$7x = 147$

$x = \frac{147}{7}$

$x = 21$

Using this let us find the measures of the smaller angle and bigger angle .

$smaller \: angle \: = x = 21°$

$larger \: angle \: = 6x + 33 = 6 \times 21 + 33 = 126 + 33 = 159°$

∴ The measure of the two angles are = 21° and 159° .

3 0

3 years ago

Find the following limit or state that it does not exist. ModifyingBelow lim With x right arrow minus 2 StartFraction 3 (2 x min

leva [86]

Answer:

-60

Step-by-step explanation:

The objective is to state whether or not the following limit exists

$\lim_{x \to -2} \frac{3(2x-1)^2 - 75}{x+2}$ .

First, we simplify the expression in the numerator of the fraction.

$3(2x-1)^2 -75 = 3(4x^2 - 4x +1) -75 = 12x^2 - 12x + 3 - 75 = 12x^2 - 12x -72$

Now, we obtain

$12(x^2-x-6) = 12(x+2)(x-3)$

and the fraction is transformed into

$\frac{3(2x-1)^2 - 75}{x+2} = \frac{12(x+2)(x-3)}{x+2} = 12 (x-3)$

Therefore, the following limit is

$\lim_{x \to -2} \frac{3(2x-1)^2 - 75}{x+2} = \lim_{x \to -2} 12(x-3) = 12 \lim_{x \to -2} (x-3)$

You can plug in $-2$ in the equation, hence

$12 \lim_{x \to -2} (x-3) = 12 (-2-3) = -60$

6 0

4 years ago

A composition that involves a translation and a reflection is called a ____________.

Olegator [25]

I think it is a glide reflection.

8 0

3 years ago