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4 years ago

Help me find the arc length of this. Thank you.

Mathematics

1 answer:

BabaBlast [244]4 years ago

4 0

The red chord is a diameter of the circle, so the length of the blue arc is half the circle's circumference.

The circumference of a circle with diameter $d$ is $\pi d$ , so the circle has circumference $8\pi$ , and so the arc has length $4\pi$ .

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Answer:

-4.416 repeating

Step-by-step explanation:

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3 years ago

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The box plots shown represent two data sets. Use the box plots to compare the data sets.

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Answer:

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3 years ago

What is 6/15 * 8 equal

Alona [7]

Answer:

16/5 or 3 1/5 or 3.2

Step-by-step explanation:

6/15*8

Dividing by 3 on both numerator and denominator,

2/5 *8 = (2*8)/5 = 16/5

6 0

3 years ago

A piece of wire 19 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

mr Goodwill [35]

Answer: 8.26 m

Step-by-step explanation:

$Let s be the length of the wire used for the square. \\Let $t$ be the length of the wire used for the triangle. \\Let $A_{S}$ be the area of the square. \\Let ${A}_{T}}$ be the area of the triangle. \\One side of the square is $\frac{s}{4}$ \\Therefore,we know that,$$A_{S}=\left(\frac{s}{4}\right)^{2}=\frac{s^{2}}{16}$

$The formula for the area of an equilateral triangle is, $A=\frac{\sqrt{3}}{4} a^{2}$ where $a$ is the length of one side,And one side of our triangle is $\frac{t}{3}$So,We know that,$$A_{T}=\frac{\sqrt{3}}{4}\left(\frac{t}{3}\right)^{2}$$We have to find the value of "s" such that,$\mathrm{s}+\mathrm{t}=19$ hence, $\mathrm{t}=19-\mathrm{s}$And$$A_{S}+A_{T}=A_{S+T}$

$$$Therefore,$$\begin{aligned}&A_{T}=\frac{\sqrt{3}}{4}\left(\frac{(19-s)}{3}\right)^{2}=\frac{\sqrt{3}(19-s)^{2}}{36} \\&A_{T+S}=\frac{s^{2}}{16}+\frac{\sqrt{3}(19-s)^{2}}{36}\end{aligned}$

$Differentiating the above equation with respect to s we get,$$A^{\prime}{ }_{T+S}=\frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}$$Now we solve $A_{S+T}^{\prime}=0$$$\begin{aligned}&\Rightarrow \frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}=0 \\&\Rightarrow \frac{s}{8}=\frac{\sqrt{3}(19-s)}{18}\end{aligned}$$Cross multiply,$$\begin{aligned}&18 s=8 \sqrt{3}(19-s) \\&18 s=152 \sqrt{3}-8 \sqrt{3} s \\&(18+8 \sqrt{3}) s=152 \sqrt{3} \\&s=\frac{152 \sqrt{3}}{(18+8 \sqrt{3})} \approx 8.26\end{aligned}$

$The domain of $s$ is $[0,19]$.So the endpoints are 0 and 19$$\begin{aligned}&A_{T+S}(0)=\frac{0^{2}}{16}+\frac{\sqrt{3}(19-0)^{2}}{36} \approx 17.36 \\&A_{T+S}(8.26)=\frac{8.26^{2}}{16}+\frac{\sqrt{3}(19-8.26)^{2}}{36} \approx 9.81 \\&A_{T+S}(19)=\frac{19^{2}}{16}+\frac{\sqrt{3}(19-19)^{2}}{36}=22.56\end{aligned}$

$$$Therefore, for the minimum area, $8.26 \mathrm{~m}$ should be used for the square$

8 0

2 years ago

The length of a rectangular field is 6 metres longer than its width. If the area of the field is 72 square metres, What are the

True [87]

Answer:

Let's call the length of the field "l", and the width of the field "w".

If the area of the field is 72 square meters, then we have:

l x w = 72

And if the length is 6 meters longer than the width, we have:

l = w+6

So looking at the first equation (l x w = 72), we can substitute the l for a w+6.

And we obtain:

(w+6) x (w) = 72

Which simplifies to w^2 + 6w = 72.

This quadratic equation is pretty easy to solve, you just need to factor it.

w^2 + 6w - 72 = 0

(w-6)(w+12)

This leaves the roots of the quadratic equation to be 6 and -12, but in this case, a width of -12 wouldn't make sense.

So, the width of the rectangular field is 6, and the length of the field is 12.

Let me know if this helps!

5 0

3 years ago