Melody

This question appears to be very simple, have you tried drawing the diagram?

Let ABCD be a parallelogram. If AC = 9 and BD = 17, then find AB^2 + BC^2.

\(BC^2=4.5^2+8.5^2-(2*4.5*8.5*cos\alpha) \qquad (1)\\~\\ AB^2=4.5^2+8.5^2-(2*4.5*8.5*cos(180-\alpha))\\ AB^2=4.5^2+8.5^2+(2*4.5*8.5*cos(\alpha))\qquad (2)\\ (1)+(2)=\)

You can check and finish it 

30000 * 4/5 * 0.9 *0.9 *0.9

y=sqrt(x^2 + 3 - 8x + 3x^3)

I grahped it but it doesn't help a whole lot

Ty it

odd+ odd = even   (which is a multiple of 2) so we cant have 2 odd numbers

even+even=even       so we can't have 2 odd numbers

odd + even = ?    

you can think through the rest of it.

Please proofread your question.

Find the derivative using the derivatives formula, [f(x+h)-f(x)]/h of f(x)= 3x^3 + 2x^2 + x + 1

f(x)= 3x^3 + 2x^2 + x + 1

f(x+h)= 3x(x+h)^3 + 2(x+h)^2 + (x+h) + 1

\(\displaystyle \frac{f(x+h)-f(x)}{h} = \frac{[  3x(x+h)^3 + 2(x+h)^2 + (x+h) + 1] - [3x^3 + 2x^2 + x + 1] }{h}\)

now simplify

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This has been edited in order to correct copy errors.

Thanks Alan.

Very glad to hear it     

Hi Jenny

1) A tennis ball is shot vertically to a height of 20 feet and aloowed to drop. Each bounce is 80% as high as the previous bounce. What is the total distance the ball travels from when it is shot unit it bounces the fourth time?

0.8*20=16

0.8*16=12.8  

etc

so 

Up distance = 20+16+12.8+.....    sum to infinity

down distance = 20+16+12.8+ ....sum to infinity

add them together

2) Suppose you drop a tennis ball from a height of 15 feet. After the ball hits the floor, it rebounds to 85% of its previous height. What is the total distance the ball traveled.

15*0.85=12.75

12.65*0.85=10.8375

down distance = 15+12.75+10.8375+ ... sum to infinity

up distance = 12.75+10.8375+.... sum to infinity

add them together

use GP sum of infinite series to get the first answers.

Is that enough help?

You are very welcome :)