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larisa86 [58]
3 years ago
12

Write the point-slope form of the equation with the given characteristics.

Mathematics
1 answer:
Neporo4naja [7]3 years ago
3 0

Answer:

1. y = 2x+3

2. y= -2x-1

3. y=3/5 x +3

4. y = 1/2 x 23/2

Step-by-step explanation:

1. Use slope intercept, y=mx+b where m =2 and b=3. y=2x+3

2. Use the point-slope form to write the equation, then simplify and convert into the slope intercept form.  

(y-5)=-2(x--3)  

y-5=-2(x+3)

y=-2x-6+5

y=-2x-1

3. Use the point-slope form to write the equation, then simplify and convert into the slope intercept form. The slope is 3/5 since parallel lines have the same slope.  

(y-3)=3/5(x-0)  

y-3=3/5 x

y=3/5 x +3

4. Use the point-slope form to write the equation, then simplify and convert into the slope intercept form. The slope is 1/2 since perpendicular lines have the negative reciprocal slopes. So -2 becomes 1/2.

(y-10)=1/2(x--3)  

y-10=1/2 (x+3)

y-10 = 1/2 x + 3/2

y = 1/2 x 23/2



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7- a. What is the radius of the circle with center (3,10) that passes through (12,12)?
loris [4]

Answer: a. Radius of circle = \sqrt{85}

b. The equation of this circle : (x-3)^2+(y-10)^2=85

Step-by-step explanation:

Given : Center of the circle = (3,10)

Circle is passing through (12,12).

a. To find the radius we apply distance formula (∵ Radius is the distance from center to any point ion the circle.)

Radius of circle = \sqrt{(12-3)^2+(12-10)^2}

Radius of circle = \sqrt{(9)^2+(2)^2}=\sqrt{81+4}=\sqrt{85}

i.e. Radius of circle = \sqrt{85}

b. Equation of a circle = (x-h)^2+(y-k)^2=r^2 , where (h,k)=Center and r=radius of the circle.

Put the values of (h,k)= (3,10) and r= \sqrt{85} , we get

(x-3)^2+(y-10)^2=(\sqrt{85})^2

(x-3)^2+(y-10)^2=85

∴  The equation of this circle :(x-3)^2+(y-10)^2=85

6 0
3 years ago
(6ab-8a+8) - (7ab-1 )
PSYCHO15rus [73]

Answer:

- ab - 8a + 9

Step-by-step explanation:

(6ab - 8a + 8) - (7ab - 1) \\ 6ab - 8a + 8 - 7ab + 1 \\  - ab - 8a + 9

hope this helps you.

4 0
3 years ago
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If x=3 calculate the value of x3-x
Dmitrij [34]
x^3-x=3^3-3= \\\\ =27-3= \\\\ =\boxed{24}
6 0
3 years ago
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Answer:

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4 0
3 years ago
The graph below shows the solution to which system of inequalities?
nlexa [21]

Answer:

Option D.

Step-by-step explanation:

Consider option D. y\leq \frac{3x}{4}+10\,,\,y\leq \frac{-x}{2}-3

Take point (0,0)

On putting this point in inequation y\leq \frac{3x}{4}+10 , we get

0\leq 10 which is true . So, solution is region towards the origin i,e region below the line y= \frac{3x}{4}+10 including the line itself .

On putting (0,0) in inequation y\leq \frac{-x}{2}-3 , we get 0\leq -3 which is false , so solution is region away from the origin i.e region below line y= \frac{-x}{2}-3 including the line itself .

So, common solution to both the inequations is the shaded part in the given figure .

In other words, we can say that the graph shown in the given figure represents system of equations: y\leq \frac{3x}{4}+10\,,\,y\leq \frac{-x}{2}-3

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3 years ago
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