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2 years ago

Which is an equation in point-slope form for the given point and slope?

Mathematics

1 answer:

Kisachek [45]2 years ago

8 0

The point-slope form: $y-y_1=m(x-x_1)$

Q2.

$(1,\ -7)\to x_1=1,\ y_1=-7\\m=1\\\\y-(-7)=1(x-1)\\\\y+7=(x-1)$

Q3.

$(5,\ 9)\to x_1=5,\ y_1=9\\m=2\\\\y-9=2(x-5)$

Q4.

$y=mx+b\\\\(0,\ 1)\to b=1\\\\y=mx+1$

The formula of a slope: $m=\dfrac{x_2-x_1}{y_2-y_1}$

$((-2,\ 4)\to x_1=-2,\ y=4\\(2,\ -2)\to x_2=2,\ y_2=-2\\\\m=\dfrac{-2-4}{2-(-2)}=\dfrac{-6}{4}=-1.5$

$y=-1.5x+1$

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3 years ago

Read 2 more answers

X-4(3-2x) = 3(3x+4) Find X

Mandarinka [93]

Answer:

all x are the solution

Step-by-step explanation:

x - 4×3 - 4×(-2x) = 3×3x + 3×4

x - 12 + 8x = 9x +12

9x + 12 = 9x + 12

9x = 9x

(0 = 0)

no matter which number we take for x the equation will always be correct

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2 years ago

What is the solution to this system of linear equations? x + y = 4 x − y = 6

Stells [14]

Answer:

x=5 , y= -1

Step-by-step explanation:

You can see that one of the y's is negative while the other is positive, so you can add them together.

Now, negative 1 + positive 1 = 0, so the y's cancel out leaving only the x's.

Remember you can't add one part without adding the other so x+x = 2x, and 4+6=10. This makes the final equation 2x=10.

You can simplify that into x=5. This gives you the answer for x

Then substitute x=5 into any of the two equations (doesn't matter) I'll use the first one. So if you substitute it you get 5+y=4.

Simplified, y= -1

Therefore, x= 5 and y=-1

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2 years ago

A rectangular piece of land to be developed as a memorial park has an area of 75 m^2. The length of the lot is three times the w

Lelechka [254]

Let length = l

width = w

length = 3 times width = 3w

Area of the land = 75 sq. m

Area = length × width = 3w×w = 75

3w²=75

w²=25

w=5 m

length = 3×15 = 15m

Now lets make the figure with the path of width x meters

(Refer the attached figure for easy understanding)

The inner length is the length of the rectangle reduced by x meters on either side due to the path

that is 15-x-x = 15-2x

the inner width is the width reduced by x meters in either side

= 5-x-x = 5-2x

Now lets calculate the inner area = inner length × inner width

(15-2x)×(5-2x)

= $75-30x-10x+4x^{2}$

= $75-40x+4x^{2}$ sq. m

Now the function to calculate the cost

f(x) is given by $3.00 × area of the inner land

f(x) = 3 × {75-40x+4x²)

= 225-120x+12x²

= 12x²-120x+225

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2 years ago

What is the area of a wall that is 28' wide by 10' high at the sides and has a triangular gable that measures 7' high?

Karolina [17]

Answer:

378 ft²

Step-by-step explanation:

Given:

The Width of the wall = 28'

The Length of the wall = 10'

The Height of the gable = 7'

Now, the width of the gable will be equal to the width of the wall = 28'

therefore,

Area of the wall = Area of the rectangle section of wall + area of the gable

Total area of the wall = ( 28' × 10' ) + ( $\frac{1}{2}\times28'\times7'$ )

Total area of the wall = 280 + 98 = 378 ft²

7 0

2 years ago