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3 years ago

HELP ME PLS Solve x-3y-6=0

Mathematics

1 answer:

Ne4ueva [31]3 years ago

7 0

Answer: x=3y+6
Explanation to solve:
You must move all terms not containing x to the right side of the equation!

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Keisha went into a park at 1:30 pm she hiked for 1 hour 35 minutes. Then she went to the picnic area for 45 minutes and left the

nata0808 [166]

Answer:

3:50pm

Time Keisha went to the park- 1:30pm

Time spent for hiking- 1hour 35 minutes

Time spent at the picnic area- 45 minutes

1: 30pm

+1. 35min

2: 65pm

But 60 minutes is 1 hour so the 65 will be changed to 1 hour 5 minutes making it

3: 05pm

+0: 45min

3: 50pm

Hence the time Keisha left the park was 3:50pm

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3 years ago

Let L: R2 --> R2 be a linear operator.If L((1,2)T) = (-2,3)Tand L((1,-1)T) = (5,2)TFind the value of L((7,5)T

devlian [24]

Answer: L((7,5)T)=(7, 18)T

Step-by-step explanation:

The step by step explanation is given in picture.

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3 years ago

Kyle is painting the front door of his house. The dimensions of the door are 80 inches by 36 inches by 2 inches. If he paints al

kolbaska11 [484]

Answer:

its 5760ins

Step-by-step explanation:

80 inches * 36 inches * 2 inches = 5760 inches3

good day, and be safe

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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

Solve D=RT for D, if T=5hours and R=65mph

brilliants [131]

D=325. u multilply 5 times 65 and thats the answer u get

4 0

3 years ago