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MariettaO [177]
3 years ago
15

The length of the outside stage is 6 feet less than the width.if the total area of the outdoor stage is 720ft^2 what is the widt

h of the stage ?
Mathematics
1 answer:
Vladimir [108]3 years ago
4 0

Answer: 30


Step-by-step explanation:

you should draw a model of this. Draw a rectangle labeling opposite sides as x (width). then label the other 2 sides as x-6 (which is the length). If x(x-6)=720, then you should be able to solve. x^2 (squared)-6x=720. Then make it into a quadratic. x^2-6x-720, and solve. You'll get (x-30) and (x+24) meaning x can either equal 30, or -24. If you didn't know- you get positive 30 because x-30=0 (that's what (x-30) and (x+24) means). Obviously a side length can't be negative, so it's 30

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kondaur [170]

Answer:

7/9 and 6

Step-by-step explanation:

Coefficients are intergers that are parted with a variable.

7/9 and 6 are the coefficients

7 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
Little Snail is going to visit his friend over at the next pond, 3 miles away. He can crawl ( 1/2. 1/3, 1/4, 3/4, 2/3 ) of a mil
KIM [24]

Solution :

It is given that Little Snail is going to visit one of his friend at the pond which is 3 miles away.

When the snail crawls 1/2 of a mile per day, it will take him, $1 \times \frac{2}{1} \times 3$

   =  6 days to get to the next pond.

When the snail crawls 1/3 of a mile per day, it will take him, $1 \times \frac{3}{1} \times 3$

   =  9 days to get to the next pond.

When the snail crawls 1/4 of a mile per day, it will take him, $1 \times \frac{4}{1} \times 3$

   = 12 days to get to the next pond.

When the snail crawls 3/4 of a mile per day, it will take him, $1 \times \frac{4}{3} \times 3$

   =  4 days to get to the next pond.

When the snail crawls 2/3 of a mile per day, it will take him, $1 \times \frac{3}{2} \times 3$

   =  4.5 days to get to the next pond.

3 0
3 years ago
Is y= 3(1/2)^x increasing or decreasing?
Zigmanuir [339]

Answer: Hey I looked on a graphing calculator on Desmos, and its seems that this is increasing.

Step-by-step explanation:

5 0
3 years ago
Jovie is maintaining a camp fire. She has kept the fire steadily burning for 101010 hours with 151515 logs. She wants to know ho
Eva8 [605]

Answer:666 hours

Step-by-step explanation: The reason is that if you turn the problem into an equation it would be h=Lx. h= hours. L=how long the log lasts and x=how many logs. So when you plug in the numbers you get 101010=L*151515. So we need to find L. What you do is you divide both sides by 151515 since it is the opposite of multiplication. 151515/151515 gets crossed out and 101010/151515 is .6666666666666 irrational. So the equation now looks like .666666 irrational=L. So .66666 irrational is your L. Know you plug .666666 irrational into your original equation. Which is now h=.6666 irrational*x. So to find how long the fire keeps on burning with 999 logs you just plug 999 into x and now your equation looks like this h=.6666 irrational*999. If you multiply .6666 irrational by 999 your final answer is 666.

8 0
3 years ago
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