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3 years ago

Help it’s URGENT!!!!!

Mathematics

1 answer:

umka21 [38]3 years ago

4 0

n=number of hours

inequality = 3n+10=40

Solution = n = 10. So 10 hours.

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Write an equation in slope intercept form that passes through -2, 3 and Is parallel to the line with equation 3X +2y=6

soldier1979 [14.2K]

The equation of line is: $y=-\frac{3}{2}x$ as the y-intercept is zero

Step-by-step explanation:

Given

(-2,3)

Equation of line => 3x+2y=6

Converting the equation in slope intercept form

$3x+2y=6\\2y=-3x+6\\Dividing\ both\ sides\ by\ 2\\\frac{2y}{2}=-\frac{3}{2}x+\frac{6}{2}\\y=-\frac{3}{2}x+3$

The co-efiicient of x is the slope of the fiven line

So,

m=-3/2

The line parallel to given line will have the same slope as slopes of two parallel lines are equal

Now,

Slope-intercept form is:

$y=mx+b$

Putting the value of slope

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

to find the value of b, put (-2,3) in the equation

$3=-\frac{3}{2}(-2)+b\\3=3+b\\b=0\\$

Putting the value of b and m

$y=-\frac{3}{2}x+0\\y=-\frac{3}{2}x$

Hence,

The equation of line is: $y=-\frac{3}{2}x$ as the y-intercept is zero

Keywords: Slope-intercept form, Slope

Learn more about slope-intercept form at:

#LearnwithBrainly

4 0

3 years ago

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago

CAN SOMEONE HELP ME ON MY WORK PLEASE :( it’s on exponential functions

strojnjashka [21]

Answer:

what's the math problem

Step-by-step explanation:

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3 years ago

What is the answer to negative 1/9 times 6 plus 5 equal.

Naddik [55]

Answer: 17/3

Decimal Form Answer: 5.666667

Step-by-Step:

Add 2/3+5 and get your answer 17/3

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3 years ago

Read 2 more answers

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Alexxandr [17]

Answer:

Nice!!

Step-by-step explanation:

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3 years ago