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Mademuasel [1]
4 years ago
8

Write an equation for the line that is parallel to the given line and passes through the given point y=5x+10;(2,14)

Mathematics
2 answers:
PilotLPTM [1.2K]4 years ago
4 0

if the lines are parallel the slopes are the same so m=5

the equation of a line given the slope and a point is

y-y1 = m(x-x1)

y-14= 5(x-2)

y-14 = 5x-10

y=5x+4

Ede4ka [16]4 years ago
3 0

Answer:

21

Step-by-step explanation:

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Lady Gaga bought a package of pens for $2.25 and some pencils that cost $.30 each she paid a total of $4.65 for these items befo
Alexxx [7]

Answer:

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Step-by-step explanation:

i used a calculator so i am 100% sure these are right.

7 0
3 years ago
What is the range of g?
Alexandra [31]

Answer:

  [-4, 9]

Step-by-step explanation:

The range is the vertical extent of the relation.

__

The minimum value of g is marked by the solid dot at (x, y) = (4, -4). That minimum is -4.

The maximum value of g is the peak at (x, y) = (-2, 9). That maximum is 9.

The function is continuous between these values, so the range includes all values between -4 and +9, inclusive.

  range: [-4, 9]

3 0
3 years ago
What is the value of b that makes the lines, given by 3x-y=4 and 2y+4bx=1, perpendicular?
USPshnik [31]

b = 1/6

Explanation:

3x-y=4 ...1st equation

Rewriting in slope- intercept form:

y = 3x - 4

2y+4bx=1

Rewriting in slope- intercept form:

2y = -4bx + 1

y = -4bx/2 + 1/2

Equation of line: y = mx + c

where m = slope, c = intercept

For a line to be perpendicular to another, the slope of one will be the negative reciprocal of the slope of another.

Slope of the 1st equation = 3

m = 3

reciprocal of 3 = 1/3

negative reciprocal of 3 = -1/3

Slope of the 2nd equation = -4b/2

we equate both slope:

negative reciprocal of 3 = -4b/2

-1/3 = -4b/2

cross multiply:

-1(2) = 3(-4b)

-2 = -12b

Divide both sides by -12:

-2/-12 = -12b/-12

b = 1/6

Hence, the value of b that makes the lines perpendicular is 1/6

6 0
1 year ago
The angle θ lies in Quadrant II .
Andreyy89

let's keep in mind that, in the II Quadrant, cosine is negative and sine, is positive.

cosine is adjacent/hypotenuse, however the hypotenuse is simply a radius unit, and thus is never negative, so in the -(2/3) the negative must be the numerator, -2.


\bf cos(\theta )=\cfrac{\stackrel{adjacent}{-2}}{\stackrel{hypotenuse}{3}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{3^2-(-2)^2}=b\implies \pm\sqrt{5}=b\implies \stackrel{\textit{II Quadrant}}{+\sqrt{5}=b}~\hfill tan(\theta )=\cfrac{\stackrel{opposite}{\sqrt{5}}}{\stackrel{adjacent}{-2}} \\\\\\ ~\hspace{34em}

6 0
4 years ago
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