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2 years ago

Write an equation in point-slope form of the line that passes through the given point and has the given slope. (12, 5); m = -3

Mathematics

1 answer:

givi [52]2 years ago

7 0

Answer:

y - 5 = -3(x - 12)

Step-by-step explanation:

Point slope form is: y - y1 = m(x - x1), where (x1, y1) is a point on the line, and m is the slope of the line.

Plug in the point and the slope:

y - y1 = m(x - x1)

y - 5 = -3(x - 12)

So, y - 5 = -3(x - 12) is the equation in point slope form of the line.

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One linear expression is subtractive from a second linear expression and the difference is X -5 what is the difference when the

lesya692 [45]

For this case, the first thing we must do is identify the linear expressions.
first linear expression = y1 = x
second linear expression = y2 = 5
We have then that the difference of the second with respect to the first is:
y2-y1 = 5-x
Answer:
The difference when the second line is here is subtracted from the first is:
5-x

7 0

3 years ago

What are the amplitude, period, and phase shift of the given function? f(t)=-2/3 cos (3t-3pi)

dybincka [34]

Answer:

amplitude; $\frac{2}{3}$

Phase shift; $\pi$ units right

Period; $\frac{2\pi}{3}$

Step-by-step explanation:

The given function is

$y=-\frac{2}{3}\cos(3t-3\pi)$

This function is of the form;

$y=A\sin (Bt+C)$

The period is given by:

$|A|=|-\frac{2}{3}|= \frac{2}{3}$

The period is given by:

$T=\frac{2\pi}{|B|}= \frac{2\pi}{|3|}=\frac{2\pi}{3}$

The phase shift is given by;

$\frac{C}{B}=\frac{-\3pi}{3}=- \pi$ or $\pi$ units right.

4 0

2 years ago

On a map, the scale shown is 1 inch : 5 miles. If a wildlife refuge is 50 square miles, what is the area of the refuge on the ma

Mama L [17]

Answer:

dababy

Step-by-step explanation:

lets go

8 0

2 years ago

ΔCAR has coordinates C (2, 4), A (1, 1), and R (3, 0). A translation maps point C to C' (3, 2). Find the coordinates of A' and R

shutvik [7]

Answer:

$A'= (2,-1)$ and $R'=(4,-2)$ under this translation.

Step-by-step explanation:

A translation in $R^{2}$ is a mapping T from $R^{2}$ to $R^{2}$ defined by $T(x,y) = (x + v_1,y+v_2)$ , where $v=(v_1,v_2)$ is a fixed vector in $R^{2}$ .

From the problem we know that $T(2,4)=(3,2)$ , so we need to find the values $v_1$ and $v_2$ such that $T(2,4) = (2 + v_1,4+v_2)=(3,2)$ , so $3=2 + v_1$ and $4+v_2=2$ , thus $v_1=1$ and $v_2=2$ .