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3 years ago

Find the equation of a line with the given points.(put your answers in the form y=mx+b.)(7,2)(5,4)

1 answer:

3 0

4-2 over(draw a line) 5-7 =2 over -2 =-1 Next find your y intercept 4=-1(5) multiply -1 ×5 = -5 add 5 +4 =9 Final answer=y=-1×+9

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Annette [7]

Answer:

The equation of line in the form of Ax + By = C is: $\mathbf{4x+y=-17}$

Step-by-step explanation:

We need to find the equation of line, having slope m = -4 and passes through point (-2,-9)

First we need to find slope and y-intercept of the line.

The slope is given i.e. m = -4

Now, we need to find the y-intercept

Using the point (-2,-9) and slope m = -4, y-intercept can be found using:

$y=mx+b\\-9=-4(-2)+b\\-9=8+b\\b=-9-8\\b=-17$

Now, the equation of required line having slope m = -4 and y-intercept b = -17 is:

$y=mx+b\\y=-4x-17$

Now writing in form of Ax + By = C

$y=-4x-17\\4x+y=-17$

So, the equation of line in the form of Ax + By = C is: $\mathbf{4x+y=-17}$ .

5 0

2 years ago

I'm confused on this one

pishuonlain [190]

Ok, I'm only in 8th grade but I believe you do addition first in the problem. Hope this helps! <3

7 0

3 years ago

Solve for X (picture included)

Fynjy0 [20]

Answer:

x = 41

Step-by-step explanation:

(2x+20) + (2x-4) = 180 degrees

4x+16 = 180 degrees

4x = 164

x = 41

The two sides are supplementary and add up to 180 degrees.

6 0

3 years ago

a relay team of 4 people runs 3 mile. it each person runs the same distance, how many miles does each person run?

Bingel [31]

Each person runs 3/4 miles.

6 0

2 years ago

2. Consider the circle x^2−4x+y^2+10y+13=0. There are two lines tangent to this circle having a slope of 2/3.

likoan [24]

Answer:

a) The coordinates are (0.431, -2.646) and (3.568,-7.354)

b) The tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

Step-by-step explanation:

First, lets complete squares, by taking for each cordinate the square of a linear expression

0= x²−4x+y²+10y+13 = (x-2)²+ 4 + (y+5)²-25+13 = (x-2)²+(y+5)² - 8

Hence (x-2)² + (y+5)² = 8

Lets put y in function of x. We should obtain 2 functions f and g that represent the circle.

(y+5)² = 8 - (x-2)² = -x² + 4x + 4

$y = ^+_- \sqrt{-x^2+4x+4} -5$

Thus

$f(x) = \sqrt{-x^2+4x+4} - 5$

$g(x) = -\sqrt{-x^2+4x+4} - 5$

Lets find the derivate of each function and the points in which they reach the value 2/3. In those points the tangent line will have a slope of 2/3.

We may just find the values of x which the derivate of f is either 2/3 or -2/3.

$\frac{-x+2}{\sqrt{-x^2+4x+4}} = ^+_-\frac{2}{3} \Leftrightarrow \frac{x^2-4x+4}{-x^2+4x+4} = \frac{4}{9} \Leftrightarrow 9(x^2-4x+4) = 4(-x^2+4x+4) \\\Leftrightarrow 13x^2-52x+20 = 0$

The quadratic has roots

$\frac{52 ^+_-\sqrt{1664}}{26}$

one root is 3.568, which corresponds with g, and the other root is 0.431, corresponding to f.

Also g(3.568) = - 7.354 and f(0.431) = -2.646

This means that the coordinates are (0.431 , -2.646), (3.568 , -7.354) and the tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

5 0

2 years ago