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dexar [7]
3 years ago
14

I just need the Point-Slope Form equation

Mathematics
1 answer:
Alina [70]3 years ago
7 0

The point-slope equation formula is (y - y_1) = m(x - x_1), where m is the slope of the line and (x_1, y_1) is a point on the line.


In our problem, we are adding $200 for each week x, or we are changing each week by +200. This means that our slope m, which is essentially the change of the line, is 200. The problem also gave us a coordinate point on the line, which was that at week 4, or when x = 4, we have $1700 in the account, or y = 1700. Thus, the coordinate point is (4, 1700).


Using this information, we can find the equation of the line:

(y - 1700) = 200(x - 4)


The answer is (y - 1700) = 200(x - 4).

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6x+9y=15<br> -12x-18y=-30<br> Solve system by elimination <br> Giving 10,pts
larisa [96]

Answer:

dssfgffdddfggrrghygfdedfhyjjujtg

7 0
2 years ago
Find the distance between the two points. (-1, 6), (2, 8) Round to the nearest hundredth
MakcuM [25]

Answer:

d = \sqrt{13} or d = 3.61

Step-by-step explanation:

d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\d= \sqrt{(2-(-1))^2+(8-6)^2}\\d= \sqrt{(3)^2+(2)^2}\\d= \sqrt{9+4}\\d= \sqrt{13}

5 0
3 years ago
To hang lights up on his house.Garrett place is a 14 foot ladder 4 feet from the base of the house. How high up the house will t
Mariulka [41]

This seems like a right triangle problem.

So assuming 14 feet is the hypotenuse and 4 feet is a leg of the right triangle, we can use the pythagorean theorem (a^{2} +b^{2} =c^{2}) to solve for the height of the house, in which we shall name it x.

So, the equation is 4^{2} +x^{2} =14^{2}.

Solve for x:

16+x^{2}= 196

x^{2}= 196-16

x^{2}= 180

x = 6√5 feet


Hope this helps!

6 0
3 years ago
Read 2 more answers
Please answer this question too! :D
Artyom0805 [142]

Answer:

A

Step-by-step explanation:

We can find the surface area of the object by adding the surface areas of each part. We have many rectangle faces to count and two triangular faces. Each has a formula for the area. We will find the area of each and then add them all together.

Triangle - 0.5 *b*h

Rectangle - b*h

<u>Triangles</u>

There are two triangles on either side. The height is 1.5. The base is 1.8.

0.5(1.5)(1.8)=1.35 meters squared

Since there are two, we will add 1.35+1.35 in our final calculation.

<u>Rectangles</u>

We will start by calculating the largest rectangle on the side. It has height of 4 and a base of 2.5 (shown above left).

4(2.5)=10

Since there are two (one we can see and one we can't), we will add 10+10 in our final calculation.

Next we calculate the top and bottom. The height is 3 and the base is 2.5 on top. But the bottom sticks out more and adds 1.8 to its base.

Top - 3(2.5)=7.5

Bottom-3(2.5+1.8)=12.9

Finally, we will calculate the front side and back(not visible) as well as the slant up front. The back side has height 4 and base 3. The front side has base 3 and height 4-1.5=2.5. The slant has base 2.3 and height 3.

Back - 4(3)=12

Front- 3(2.5)=7.5

Slant - 3(2.3)=6.9

We add all together for the total surface area: 1.35+1.35+10+10+7.5+12.9+12+7.5+6.9=69.5 meters squared.

5 0
3 years ago
What is the cross-sectional area of a cement pipe 2 in. thick with an inner diameter of 5 ft? Use pie 3.14.
Paul [167]

Answer:

1.60 ft^{2}

Step-by-step explanation:

Assuming that this is a cylindrical pipe, the area can be determined by applying the formula for calculating the area of a circle.

i.e Area = \pir^{2}

So that,

for the inner part of the pipe,

r = \frac{diameter}{2}

 = \frac{5}{2}

 = 2.5 feet

Area of the inner part of the pipe = \pir^{2}

                                 = 3.14 x (2.5)^{2}

                                 = 19.625

Are of the inner part of the pipe is 19.63 ft^{2}.

Total diameter of the pipe = 5 + 0.2

                                           = 5.2 feet

r = \frac{5.2}{2}

 = 2.6 feet

Area of the pipe = \pir^{2}

                            = 3.14 x (2.6)^{2}

                            = 21.2264

Area of the pipe is 21.23 ft^{2}.

Thus the cross sectional area = Area of the pipe - Area of the inner part of the pipe

                                           = 21.2264 - 19.625

                                           = 1.6014

The cross sectional area of the cement pipe is 1.60 ft^{2}.

6 0
2 years ago
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