Y=[? ]°<br><br> Please help me!!

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

3 years ago

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AB is a directed segment with endpoints A(2,2) and B(14,14). In two or more complete sentences, explain how you would verify tha

GenaCL600 [577]

The smaller side is 4 units going from point (2,2) to point (6,6) while the biger side is 8 units long going from point (6,6) to point (14,14). So 4 is 1/2 of 8 therefore there is a 1:2 ratio because of 4:8

5 0

3 years ago

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A random sample of 102 full-grown lobsters had a mean weight of 16 ounces and a standard deviation of 3.4 ounces. Construct a 98

pshichka [43]

Answer:

The best point estimate for a confidence interval estimating the population μ is 16 ounces.

The 98 percent confidence interval for the population mean μ is between 15.21 ounces and 16.79 ounces.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The best point estimate for a confidence interval estimating the population μ is

The sample mean, so 16 ounces.

T interval

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 102 - 1 = 101

98% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 101 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.98}{2} = 0.99$ . So we have T = 2.36

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.36\frac{3.4}{\sqrt{102}} = 0.79$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 16 - 0.79 = 15.21 ounces

The upper end of the interval is the sample mean added to M. So it is 16 + 0.79 = 16.79 ounces

The 98 percent confidence interval for the population mean μ is between 15.21 ounces and 16.79 ounces.

4 0

3 years ago

Sally can paint a room in 8 hours while it takes steve 7 hours to paint the same room. how long would it take them to paint the

77julia77 [94]

Rate for Sally is 1 room per 8hrs
rate for Steve is 1 room per 7hrs

combined rate:
1/8 + 1/7 = 1/T
7/56 + 8/56 = 1/T
15/56 = 1/T
56/15 = T
3 11/15 = 3 hrs 44 min

3 0

3 years ago

Greatest common factor of 33 and 66

Firdavs [7]

the answer is 11 hope this helps

brainliest

4 0

3 years ago

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Y=[? ]° Please help me!!