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3 years ago

What is the value of any number raised to the 0 power?

Mathematics

1 answer:

kirill115 [55]3 years ago

4 0

Always be 1, stated in the Zero Power property

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Solve for y.<br><br> 7y - 14 = 35

Radda [10]

Answer:

it's 7

Step-by-step explanation:

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3 years ago

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Explain box and whisker and how does it work?

bezimeni [28]

a box plot is a graph that summarizes numerical data based on quartiles, which divide a data set into fourths

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3 years ago

Points A and B are on the different sides of a line 1, the distance between point A and the line lis 10 in, the distance between

givi [52]

Answer: Distance between O and the line 1 is 3 inches.

Explanation:

Since we have given that

Distance between point A and the line 1 = 10 inches

Distance between point B and the line 1 = 4 inches

Total length of segment AB is given by

$10+4=14\ inches$

Since O is midpoint of the line segment AB.

so, AO is given by

$\frac{14}{2}=7\ in.$

Now, Distance between O and the line 1 is given by

$10-7=3\ in.$

Hence, Distance between O and the line 1 is 3 inches.

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3 years ago

SOMEONE ANSWER QUICK!!

Thepotemich [5.8K]

Answer:

3/4 pizza is left. If you want an explanation, ask me ;>

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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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