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4 years ago

Can you help me with this question please

Mathematics

1 answer:

Yuki888 [10]4 years ago

7 0

Answer: x=70

Step-by-step explanation:

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The area of a circle depends on the radius, use the equation A(r) to compute A(8), Round your answer to the nearest hundredth

vovangra [49]

Area = $\pi$ x radius²

A(8) = 3.14 x 8² = 3.14 x 64 = 200.96

6 0

4 years ago

You earn commission on a $500 couch. How much do you earn?

aleksley [76]

Answer:

When an employee earns a commission, they make a portion of the sale in income. For example, if an employee sells a couch for $500 and they get a 10% commission on all sales, then they earn $50 on that sale.

Step-by-step explanation:

So it depends on what the percentage of the commission is

3 0

3 years ago

Read 2 more answers

2. Travis buys 2 quarts of milk at the store. What would be the equivalent or same as in

Vinvika [58]

Answer:

4 pints

Step-by-step explanation:

There are 2 pints in a quart, and there are 2 quarts, so 2 x 2 is 4.

3 0

3 years ago

Keep seeing adds to upgrade why?

Ne4ueva [31]

Answer:

bc they want you to spend your money on having no ads. it's pointless tbh unless you want to get a tutor on here..

Step-by-step explanation:

7 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