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2 years ago

Which two square roots are used to estimate the square root of 67

Mathematics

1 answer:

viva [34]2 years ago

5 0

$\bf \boxed{\sqrt{64}}\rule[0.35em]{10em}{0.25pt}\sqrt{67}\rule[0.35em]{19em}{0.25pt}\boxed{\sqrt{81}}$

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Somebody pls help I’m doin a exam!!!!

ra1l [238]

Answer:

Step-by-step explanation:

12x+2y=36

2y=36-12x

5x+36-12x=22

7x=14

x=2

2y=36-12(2)

2y=12

y=6

btw,next time, please do not include words like(failing ,exam,test,etc) in your question,otherwise, the question might be deleted.

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2 years ago

What is the answer to this question?

kirza4 [7]

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3 years ago

What is 52 divided by 8

Ivanshal [37]

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3 years ago

Find the sizes of the angles marked with a letter.<br> Why is my answer wrong?

jeyben [28]

Answer:

29.28 degrees.

Step-by-step explanation:

sin x / 16.2 = sin 49 / 25

Cross multiply:

25 sin x = 16.2 * sin 49

sin x = (16.2 * sin 49) / 25

sin x = 0.48905

x = 29.28 degrees.

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

4 0

3 years ago

Read 2 more answers