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3 years ago

Help me with 11 20 points!

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

6 0

Part A- multiply it by 16
Part B- 8 flower pots

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A prism and two nets are shown below: 3 in 5 In. 8.6 In. 4 In. Prism B A A D Not A Not B Part A: Which is the correct net for th

tatuchka [14]

Answer:

Step-by-step explanation:

6 0

2 years ago

32 ft/sec to meters/min

wariber [46]

585.216 would be the answer

3 0

3 years ago

10 .find the partial sum S10 of the arithmetic series with a1 = 4, d = 5.

Crazy boy [7]

Answer:

S10 = 320

Step-by-step explanation:

The formula for calculating the sum of the nth term of an AP is expressed as

Sn = n/2(2a+(n-1)d))

Given

a = 4

d = 5

n = 10

S10 =10/2{2(5)+(10-1)5}

S10 =5{10+9(6)}

S10 =5{10+54}

S10 = 5(64)

S10 = 320

Hence the sum of the first ten terms of the sequence is 320

8 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

a boy flying a kite is standing 30ft from a point directly under the kite. if the string to the kite is 50ft long,what is the an

MaRussiya [10]

30 × 30 + 50 × 50 = 3400

square root of 3400 is 58.31 feet.

58.31 feet is the angle of elevation of the kite.

3 0

4 years ago

Read 2 more answers