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3 years ago

The area of a square can be found using the equation A = s², where A is the area and s is the measure of one side of the square.

Mathematics

2 answers:

noname [10]3 years ago

5 0

The correct answer is B. s = √81 in. Hope I helped!

algol133 years ago

5 0

Answer: S= $\sqrt{81}$ in.

Step-by-step explanation: <u><em>I did the test</em></u>

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9r-5r+2+r ????????? Pls help me

Oksanka [162]

Answer:

5r + 2

Step-by-step explanation:

9r-5r+2+r=5r+2

\mathrm{Group\:like\:terms}

=9r-5r+r+2

\mathrm{Add\:similar\:elements:}\:9r-5r+r=5r

=5r+2

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3 years ago

Does a trend line always pass through every point on a scatter plot?

forsale [732]

No, it does not and it's not supposed to. A trend line is drawn to go through in the best way possible, to go through the "center" of the points as best as it can. It may hit some of the points, but not all of them. Generally, the idea is to try to get the trend line to have as many points "above" it as it does "below" it to say you've gone through them evenly.

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3 years ago

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35percent as a fraction in simplest form

Luden [163]

Answer:

7/20

Step-by-step explanation:

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3 years ago

Help with math please

JulijaS [17]

You can first multiply your coefficients (7x9), which gives you 63. You can then simplify the exponents by using the rule that when two terms with the same variable are being multiplied, you can simplify by adding exponents. Since 7+3= 10, your simplified answer would be 63x^10 :) hope it helped!

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4 years ago

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

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4 years ago

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