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iris [78.8K]
3 years ago
6

The area of a rectangular patio is 558 square yards, and its length is 112 yards. What is the patio’s width, in yards?

Mathematics
1 answer:
Paha777 [63]3 years ago
7 0
D 8716 im not 100% sure tho
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Answer:they each contributed 6 $

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1. Is the number 0.555, rational or irrational?
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Rational hope this helps
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Stuck at this question I need in 5 mins! Need it ASAP
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Answer:45



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A triangle has side lengths of 9 in, 13 in, and 20 in. What is the measurement of this triangle’s largest angle?
zvonat [6]
1. Given any triangle ABC with sides BC=a, AC=b and AB=c, the following are true :

      i) the larger the angle, the larger the side in front of it, and the other way around as well. (Sine Law) Let a=20 in, then the largest angle is angle A.

       ii) Given the measures of the sides of a triangle. Then the cosines of any of the angles can be found by the following formula:

       a^{2}=b ^{2}+c ^{2}-2bc(cosA)

2. 

20^{2}=9 ^{2}+13 ^{2}-2*9*13(cosA)

400=81+169-234(cosA)   150=-234(cosA)

cosA=150/-234= -0.641

3. m(A) = Arccos(-0.641)≈130°, 

4. Remark: We calculate Arccos with a scientific calculator or computer software unless it is one of the well known values, ex Arccos(0.5)=60°, Arccos(-0.5)=120° etc
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3 years ago
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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
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