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SashulF [63]
3 years ago
12

Use the y- and x- intercepts to write the equation of the line. y-intercept (0, -6), x-intercept (-2, 0)

Mathematics
1 answer:
garri49 [273]3 years ago
3 0

Hi there! :)

We are given 2 points. It means we can use the following formula to find the slope:

m=\frac{y2-y1}{x2-x1}

m=\frac{0-(-6)}{-2-0\\} \\\frac{0+6}{-2}\\\frac{6}{-2} \\-3

So the slope is -3

First, let's write the equation of the line in point-slope form.

y-y1=m(x-x1)

y-(-6)=-3(x-0)

y+6=-3(x+0)

y+6=-3x+0

y=-3x+0-6

y=-3x-6

Hope this helps!

~Just an emotional teen who listens to her favorite song "Try Everything"

SilentNature

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Use the distributive property to simplify -4(6-3y)
denpristay [2]

Answer:

12y-24

Step-by-step explanation:

-4(6-3y)

-24+12y

3 0
3 years ago
Find the slope of each line
yuradex [85]
The slope is -2. this is because if you look at the graph it goes down 2 units and it goes right to the once so it’ll be -2/1 which equals -2.
4 0
3 years ago
At 1:00 pm the water level in a pool is 13 inches. At 1:30pm the water level is 18 inches. At 2:30pm the water level is 28 inche
SashulF [63]
From the given question we come to know of certain number of facts and they are:
At 1:00 PM the water level of the pond was = 13 inches
At 1:30 PM the water level of the pond was = 18 inches
At 2:30 PM the water level of the pond was = 28 inches
From the above given facts we can easily find the amount of water changing every half an hour.
Amount of increase in water from 1:00PM to 1:30 PM = (18 - 13) inches
                                                                                       = 5 inches
Amount of increase in water level from 1:30PM to 2:30PM = (28 -18) inches
                                                                                             = 10 inches
From the above two deductions we can come to the conclusion the the constant rate of change in water level is 5 inches for every half an hour.
4 0
3 years ago
Read 2 more answers
Solve the system of equations.
lord [1]

Answer is B.) (-3,-3) and (2,7)

3 0
3 years ago
In the isosceles △ABC m∠ACB=120° and AD is an altitude to leg BC . What is the distance from D to base AB , if CD=4cm?
7nadin3 [17]

Correct answer is: distance from D to AB is 6cm

Solution:-

Let us assume E is the altitude drawn from D to AB.

Given that m∠ACB=120° and ABC is isosceles which means

m∠ABC=m∠BAC = \frac{180-120}{2}=30

And AC= BC

Let AC=BC=x

Then from ΔACD , cos(∠ACD) = \frac{DC}{AC} =\frac{4}{x}

Since DCB is a straight line m∠ACD+m∠ACB =180

                                              m∠ACD = 180-m∠ACB = 60

Hence cos(60)=\frac{4}{x}

          x=\frac{4}{cos60}= 8

Now let us consider ΔBDE, sin(∠DBE) = \frac{DE}{DB} =\frac{DE}{DA+AB} = \frac{DE}{4+8}

DE = 12sin(30) = 6cm

7 0
3 years ago
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