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stealth61 [152]
3 years ago
9

Olivia deposit 6952 in a savings account paying 3.72% interest to the nearest dollar how much money does Olivia have in total af

ter 16 years
Mathematics
1 answer:
lianna [129]3 years ago
4 0
Pretty sure 11,090!............................
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What is the area of each piece of paper that Jodi is cutting out
Reptile [31]
The simplest (and most commonly used) area calculations are for squares and rectangles. To find the areaof a rectangle multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself tofind the area.
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A circle has a radius of 12. What is the diameter?
AveGali [126]

Answer:

24.

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the diameter is twice the radius

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9 3/4 x 8 cummutative
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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 }) }

4 0
3 years ago
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Answer this question plz
olya-2409 [2.1K]

Answer: velocity-time graph

Step-by-step explanation: The principle is that the slope of the line on a velocity-time graph reveals useful information about the acceleration of the object. If the acceleration is zero, then the slope is zero (i.e., a horizontal line). If the acceleration is positive, then the slope is positive (i.e., an upward sloping line).

8 0
3 years ago
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