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Alex_Xolod [135]
3 years ago
12

A line with a slope of -1 passes through the point (0,0). What is its equation in slope-intercept form?​ Please answer quickly.

Mathematics
1 answer:
rusak2 [61]3 years ago
4 0

Answer:

y = -x

Step-by-step explanation:

The line has a slope of -1 and it passes through point (0,0)

Taking another point (x,y) on the line;

Slope = change in y ÷ change in x

For our case; slope = \frac{y - 0}{x - 0} = -1

Cross multiplying this gives;

y = -x

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The 11th triangular number = 66
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Find the point on the graph of the given function at which the slope of the tangent line is the given slope.
mezya [45]

Answer:

A ( -2 , 9 )

Step-by-step explanation:

<u>Idea:</u> You find the first derivative of f(x), and then set it equal to the desired slope. You'll find some x. That we will use to find the point.

f'(x) = 3x^2 + 12x + 20

f'(x) = 8

3x^2 + 12x + 20 = 8

3x^2 + 12x + 12 = 0

3 ( x^2 + 4x + 4 ) = 0

3 ( x + 2 )^2 = 0

x + 2 = 0

x = -2

So, the desired point is:

A ( -2, f(-2) ) --> A ( -2 , 9 )

3 0
3 years ago
Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP
erik [133]

Answer:

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

Step-by-step explanation:

we have

y=-x^{2} +2x+10 ----> equation A

y=x+2 ----> equation B

Solve the system by substitution

substitute equation B in equation A

x+2=-x^{2} +2x+10

solve for x

-x^{2} +2x+10-x-2=0

-x^{2} +x+8=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-x^{2} +x+8=0

so

a=-1\\b=1\\c=8

substitute in the formula

x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}

x=\frac{-1\pm\sqrt{33}} {-2}

x=\frac{-1+\sqrt{33}} {-2}  -----> x=\frac{1-\sqrt{33}} {2}  

x=\frac{-1-\sqrt{33}} {-2}  -----> x=\frac{1+\sqrt{33}} {2}  

<em>Find the values of y</em>

For x=\frac{1-\sqrt{33}} {2}  

y=x+2

y=\frac{1-\sqrt{33}} {2}+2  ---->y=\frac{5-\sqrt{33}} {2}  

For x=\frac{1+\sqrt{33}} {2}  

y=x+2

y=\frac{1+\sqrt{33}} {2}+2  ---->y=\frac{5+\sqrt{33}} {2}  

therefore

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

5 0
3 years ago
Ms.Ella wants to rent a car,it cost $0.9 per mile plus $200 initial fees.If she rent the car for x miles write an expression tha
Hatshy [7]

Answer:

0.9x + 200 = cost

4 0
2 years ago
Read 2 more answers
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