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3 years ago

What is the equation of the line -3 and y intercept of 8

Mathematics

1 answer:

Anton [14]3 years ago

7 0

9514 1404 393

Answer:

y = -3x +8

Step-by-step explanation:

The slope-intercept form of the equation of a line is ...

y = mx + b . . . . . . . . line with slope m and y-intercept b

Assuming you want your line to have a slope of -3 and a y-intercept of 8, you would put -3 where m is, and put 8 where b is. This gives ...

y = -3x +8

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I need help with this one math question please!

den301095 [7]

Answer:

see explanation

Step-by-step explanation:

Assuming the fractions are being multiplied

Factorise the denominators of both fractions

x² - 3x - 10 = (x - 5)(x + 2)

x² + x - 12 = (x + 4)(x - 3)

The product can now be expressed as

$\frac{x-5}{(x-5)(x+2)}$ × $\frac{x+2}{(x+4)(x-3)}$

Cancel (x - 5) and (x + 2) on the numerators/ denominators, leaving

$\frac{1}{(x+4)(x-3)}$ = $\frac{1}{x^2+x-12}$

3 0

3 years ago

I have been having trouble doing this problem 5x + 34

Andru [333]

Answer:

Hey! Well you cant simplify it anymore.

Step-by-step explanation:

You cant add an unknown number to another number. So its still 5x+34

I Hope this helps :|

3 0

2 years ago

Fifteen more than half a number is 9

rusak2 [61]

For this case, the first thing we must do is define a variable.

We have then:

x: unknown number

We now write the equation that models the problem:

$(\frac{1}{2}) x + 15 = 9$

From here, we clear the value of x.

We multiply both sides of the equation by 2:

$(\frac{2}{2}) x + 2 (15) = 2 (9)\\x + 30 = 18$

We subtract 30 on both sides of the equation:

$x + 30 - 30 = 18 - 30\\x = -12$

Answer:

The value of the unknown number is given by:

$x = -12$

6 0

3 years ago

Read 2 more answers

Find the distance from P to l.

ki77a [65]

$Equation\ of\ a\ line\ from\ 2\ points (x_1;\ y_1);\ (x_2;\ y_2):\\\\y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\(-4;\ 2)\ and\ (3;-5)\\subtitute:\\\\y-2=\dfrac{-5-2}{3-(-4)}\cdot(x-(-4))\\\\y-2=\dfrac{-7}{7}\cdot(x+4)\\\\y-2=-(x+4)\\\\y-2=-x-4\ \ \ |add\ x\ and\ 4\ to\ both\ sides\\\\x+y+2=0\\-------------------\\$

$Point-line\ distance\\l:Ax+By+C=0;\ (x_0;\ y_0)\\\\d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}\\\\x+y+2=0\to A=1;\ B=1;\ C=2\\(1;\ 2)\to x_0=1;\ y_0=2\\subtitute\\\\d=\dfrac{|1\cdot1+1\cdot2+2|}{\sqrt{1^2+1^2}}=\dfrac{|1+2+2|}{\sqrt2}=\dfrac{|5|}{\sqrt2}=\dfrac{5}{\sqrt2}=\dfrac{5\cdot\sqrt2}{\sqrt2\cdot\sqrt2}\\\\=\boxed{\dfrac{5\sqrt2}{2}}$

4 0

3 years ago

How many solutions does 2(x-3)=10x-6-8x

Step2247 [10]

Answer: Infinitely many

Step-by-step explanation: 2(x-3)=10x-6-8x

2x-6=10x-6-8x

2x-6=2x-6

4 0

3 years ago