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2 years ago

A line passes through (4,-5) and had a slope of 5/2. Write an equation in point-slope form for this line

Mathematics

2 answers:

Arada [10]2 years ago

8 0

Answer:

y+5 = 5/2( x -4)

Step-by-step explanation:

Point slope form of a line is

( y-y1) = m(x-x1) where m is the slope and ( x1 ,y1) is a point on the line

y --5 = 5/2( x -4)

y+5 = 5/2( x -4)

Anuta_ua [19.1K]2 years ago

6 0

Answer:

y + 5 = 5/2(x - 4)

Step-by-step explanation:

Use the point slope equation: y - y1 = m(x - x1), where (x1, y1) is a known point and m is the slope.

Plug in the values:

y + 5 = 5/2(x - 4)

So, the equation is y + 5 = 5/2(x - 4)

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I'm really confused on how to do this :/ can I get some help ? :/

Sphinxa [80]

Hello!
Circular problems are the worst, am I right?
So as it says, A (Area) = $\pi$ (Pi) $r^{2}$ (Radius Squared.)
So Area = Pi times Radius Squared. The radius is the length from the center of a circle to any "edge" of the circle. So 1.5 feet is the radius.
1.5 Times 1.5 is 2.25. Pi is around 3.14. So 2.25 times 3.14. This is 7.065.
You're trying to find that red part. So you divide 7.065 by 4. It equals 1.76625.
Rounded to the nearest hundreth, it is 1.77.
Your answer is $1.77 ft^{2}$

I hope I helped!

7 0

3 years ago

A toy rocket is shot vertically into the air from a launching pad 9 feet above the ground with an initial velocity of 88 feet pe

TEA [102]

Answer:

Step-by-step explanation:

The position function for this is:

$s(t)=-16t^2+88t+9$ . We can use this equation to find the position (or height) of the rocket at ANY TIME during its flight. I could find out the height of the rocket at 3 seconds by plugging in a 3 for t and solving for s(t); I could find the height of the rocket at 12 seconds by plugging in a 12 for t and solving for s(t), etc.

The first derivative of position is velocity:

v(t) = -32t + 88.

If we are looking for the time the rocket reaches it max height, we need to remember from physics class that this happens when the velocity of the object is at 0. We set the velocity equation equal to 0 then and solve for t:

0 = -32t + 88 and

-88 = -32t so

t = 2.75 seconds. This means that 2.75 seconds after the rocket is launched, it reaches its max height. In order to find what that max height is we plug 2.75 into the position equation for t and solve:

$s(2.75)=-16(2.75)^2+88(2.75)+9$ to get that

s(2.75) = 130

The max height is 130 feet and it reaches this point at 2.75 seconds into its motion.

7 0

2 years ago

A circle passes through point (-2, -1) and its center is at (2, -1). Which equation represents the circle?

ludmilkaskok [199]

The choices are missing, but we can solve the problem from the given information

Answer:

The equation which represents the circle is (x - 2)² + (y + 1)² = 16

Step-by-step explanation:

The form of the equation of a circle is (x - h)² + (y - k)² = r², where

(h , k) are the coordinates of its center
r is the radius of it

∵ The coordinates of the center of the circle are (2 , -1)

∴ h = 2 and k = -1

The radius of the circle is the segment whose join the center of the circle and any point on the edge of the circle

∵ The circle passes through point (-2 , -1)

∴ The radius of the circle is the distance from the center (2 , -1)

and the point (-2 , -1)

The y-coordinates of the two points are equal that means the radius is a horizontal segment and its length is the difference between their x-coordinates

∵ r = 2 - (-2)

∴ r = 4

Now let us substitute h, k, and r in the form of the equation of the circle

∵ (x - 2)² + (y - -1)² = (4)²

∴ (x - 2)² + (y + 1)² = 16

The equation which represents the circle is (x - 2)² + (y + 1)² = 16

7 0

2 years ago

What is the answer to 6.59+(-7.8)-(2.41)=

strojnjashka [21]

The answer is -3.62... hope it helps! :)

6 0

3 years ago

The answer please !!

andrey2020 [161]

Answer:

-3

Step-by-step explanation:

Multiply -1 into +3, -1 x 3 = -3

3 0

2 years ago