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3 years ago

A line passes through the point (10,9) and has a slope of 3/2. Write an equation in slope intercept form for this line

Mathematics

1 answer:

Andre45 [30]3 years ago

5 0

Answer:

The equation of line passing through (10,9) and having slope 3/2 is: $y = \frac{3}{2}x-6$

Step-by-step explanation:

The slope intercept form of a line is given by:

$y=mx+b$

We are given

Slope = m = 3/2

Point = (10,9)

Putting the value of the slope in the equation we get

$y = \frac{3}{2}x+b$

b is the y-intercept. To find the y-intercept we have to put the point through which the line passes in the equation.

Putting (10,9) in the equation

$9 = \frac{3}{2}(10) +b\\9 = 15+b\\b = 9-15\\b = -6$

Putting b=-6 in the equation

$y = \frac{3}{2}x-6$

Hence,

The equation of line passing through (10,9) and having slope 3/2 is: $y = \frac{3}{2}x-6$

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Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro

Ugo [173]

Answer:

$Side\ B = 6.0$

$\alpha = 56.3$

$\theta = 93.7$

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with $\alpha, \beta, \theta$

[See Attachment for Triangle]

$A = 10cm$

$C = 12cm$

$\beta = 30$

What the question is to calculate the third length (Side B) and the other 2 angles ( $\alpha\ and\ \theta$ )

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

$B^2 = A^2 + C^2 - 2ABCos\beta$

Substitute: $A = 10$ , $C =12$ ; $\beta = 30$

$B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30$

$B^2 = 100 + 144 - 240*0.86602540378$

$B^2 = 100 + 144 - 207.846096907$

$B^2 = 36.153903093$

Take Square root of both sides

$\sqrt{B^2} = \sqrt{36.153903093}$

$B = \sqrt{36.153903093}$

$B = 6.0128115797$

$B = 6.0$ <em>(Approximated)</em>

Calculating Angle $\alpha$

$A^2 = B^2 + C^2 - 2BCCos\alpha$

Substitute: $A = 10$ , $C =12$ ; $B = 6$

$10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 180 - 144 *Cos\alpha$

Subtract 180 from both sides

$100 - 180 = 180 - 180 - 144 *Cos\alpha$

$-80 = - 144 *Cos\alpha$

Divide both sides by -144

$\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}$

$\frac{-80}{-144} = Cos\alpha$

$0.5555556 = Cos\alpha$

Take arccos of both sides

$Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)$

$Cos^{-1}(0.5555556) = \alpha$

$56.25098078 = \alpha$

$\alpha = 56.3$ <em>(Approximated)</em>

Calculating $\theta$

Sum of angles in a triangle = 180

Hence;

$\alpha + \beta + \theta = 180$

$30 + 56.3 + \theta = 180$

$86.3 + \theta = 180$

Make $\theta$ the subject of formula

$\theta = 180 - 86.3$

$\theta = 93.7$

5 0

3 years ago

Find $5,875 ÷ $6,030 rounded to the nearest tenth of a percent.

saw5 [17]

1.0% exactly 0.9743. I hope I helped!

8 0

2 years ago

Help help help help help

Sindrei [870]

DONT PRESS THE LINK ABOVE ! ITS A SCAM

3 0

2 years ago

There are two numbers such that , if both of them are individually increased by 5 and then by the same percentage as they were i

QveST [7]

Let the two numbers be x and y.
Let m = the fraction (or percentage) for increasing the numbers.

Increase x by 5, multiply (x+5) by (1+m), and set it qual to 36.
(x + 5)*(1 + m) = 36

Increase y by 5, multiply (y+5) by (1+m), and set it equal to 36.
(y+5)*(1+m) = 36

Therefore
(x+5)*(1+m) = (y+5)*(1+m)
x + 5 = y + 5
x = y
x - y = 0

Answer: 0.
The difference between the two numbers is zero.

5 0

3 years ago

Pleaz Helpz!!! I needz this done before my final!!!

Marta_Voda [28]

the perimeter of the field is 442

width=w

the length of the field is 9 yards less than quadruple W

L=4x - 9

the perimeter is P=2L+2W

you can plug in the values

P= 2(4w-9) + 2W

plug in 442 for P

442= 2(4w- 9) + 2W

442= 8w - 36 + 2W

478 =8w + 2W

478= 10W

47.8 is your answer!!!

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2 years ago