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Alenkinab [10]
3 years ago
12

choose the equation below that represents the line passing through the point (2, – 5) with a slope of −3. y = −3x − 13 y = −3x 1

1 y = −3x 13 y = −3x 1
Mathematics
2 answers:
gtnhenbr [62]3 years ago
8 0
General equation of line;
y=mx+c
y=-3x+c
Replacing for x and y;
-5=-3(2)+c
-5=-6+c
c=-5+6
c=1
y=-3x+1
Rainbow [258]3 years ago
3 0

Answer:

y = -3x + 1

Step-by-step explanation:

Let the equation of the line is y = mx + c

Where m = slope of the line

and c = y - intercept

Slope = -3

So the equation of the line will be

y = (-3)x + c

Since this line passes through a point (2, -5)

By plugging in the values in the equation,

-5 = -3(2) + c

-5 = -6 + c

c = 6 - 5

c = 1

Therefore, the equation of the line will be, y = (-3)x + 1

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Angelina_Jolie [31]

Answer: point C = (3.75, 1.5)

Step-by-step explanation:

As the direction of the distance is from A to B, we need to down the y-axis and along (to the right) the x-axis.

Find the distance between the x-coordinates of both points by subtracting the x-coordinate of A from the x-coordinate of B:

5 - 0 = 5

3/4 of the length of this distance = 0.75 x 5 = 3.75

So the x-coordinate of C will be the sum of the distance (3.75) and the x-coordinate of A (as we are "travelling" from A to B):

3.75 + 0 = 3.75

Find the distance between the y-coordinates of both points by subtracting the y-coordinate of B from the y-coordinate of A:

3 - 1 = 2

3/4 of the length of this distance = 0.75 x 2 = 1.5

So the y-coordinate of C will be the y-coordinate of A minus the distance (1.5):

3 - 1.5 = 1.5

Therefore, point C = (3.75, 1.5)

Hope that helps - i dont know what u meant by option 1,2,3 so if u have an questions or i did it wrong i will fix it <3

   

3 0
2 years ago
Find all zeros.
Jet001 [13]

Answer:

{ 0, 3, - 3 }

Step-by-step explanation:

f(x) = x(2x - 3)(x + 3)

to find the zeros let f(x) = 0 , that is

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equate each factor to zero and solve for x

x = 0

2x - 3 = 0 ⇒ 2x = 3 ⇒ x = \frac{3}{2}

x + 3 = 0 ⇒ x = - 3

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5 0
1 year ago
Rebecca was advised to invest her £12000 life savings in a special High
Evgesh-ka [11]

#a

#1st year

\\ \rm\Rrightarrow 12000+0.045(12000)=12540

#2nd year

\\ \rm\Rrightarrow 12540+0.05(12540)=13167

#3rd year

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#4th year

\\ \rm\Rrightarrow 13864.85+13864.85(0.049)=14544.2

#b

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\\ \rm\Rrightarrow 14544.2-12000=£ 2544.2

#c

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\\ \rm\Rrightarrow \dfrac{2544.2}{12000}\times 100

\\ \rm\Rrightarrow 21.2\%

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