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kotykmax [81]
3 years ago
14

The volume of a cube can be found using the equation V=S3, where V is the volume and Sis the measure of one side of the cube. Ma

tch the equation for how to solve for the side length of a cube to its description.

Mathematics
1 answer:
bulgar [2K]3 years ago
5 0

Answer:

s = ∛729 in³

Step-by-step explanation:

Since the volume is 729 in³, our equation can be plugged into:

<em>729 = s³</em>.

Isolate s by cube rooting each side: ∛729 = s.

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Twice a number deceased by 7.translate in algebraic expressions​
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Twice a number (2x) decreased by 7 (- 7).

2x - 7

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Derivative of tan(2x+3) using first principle
kodGreya [7K]
f(x)=\tan(2x+3)

The derivative is given by the limit

f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h

You have

\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h
\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}

By this identity, you have

\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}

So in the limit you get

\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h
\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}
\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}
\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}
\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}

The first two limits are both 1, and the single term in the last limit approaches 0 as h\to0, so you're left with

f'(x)=\dfrac12\sec^2(2x+3)

which agrees with the result you get from applying the chain rule.
7 0
3 years ago
Which double facts help you solve 6+5=11
Likurg_2 [28]
9+2 = 11 8+3 = 11are you going to do is just add in that's all it does
6 0
3 years ago
I know this is super easy! My brain just cannot wrap my head around my maths today. Please help soon.
zysi [14]

Hello there! the answers to your question are:

a. $15x + $20y = $100; x + y = 6.

b. Yes, he can, since the cost of 4 shirts is $60, and two pants is $40, which adds up to be how much he is spending, or $100.  

Let's start by working to solve part a. We need to set up a system of equations that models the situation given. To do this, let's set up an equation that models how many Sean will buy in total and how much he will spend.

The question says that he wants to spend $100 total, and shirts cost $15 and pants cost $20. We can model this as $15x + $20y = $100.

The question also says he wants to purchase a total of 6 items for his wardrobe, so the number of items he buys has to add up to 6. This can be modeled as x + y = 6.

This gives us our answer for part a. Now let's solve part b. We are being asked if Sean can buy 2 pants and 4 shirts. This fits one of our equations, x + y = 6, since 4 and 2 add up to be 6. We need to see if this fits our other equation, though $15x + $20y = $100. To do this, we can substitute the number of shirts (4) into the x value, and the number of pants (2) into the y value and solve. If the total is more than 100, he can't. If it is less, he can.

$15(4) + $20(2) = 100

60 + $20(2) = 100

60 + 40 = 100

Since our total adds up to $100, he can buy 2 pairs of pants and 4 shirts.

I hope this helps and have a great rest of your day!

8 0
3 years ago
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