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Dennis_Churaev [7]
3 years ago
12

I need help with this!!!!

Mathematics
2 answers:
Lady bird [3.3K]3 years ago
6 0

Answer:

Tomas is the fastest

Step-by-step explanation:

To find out who is the fastest, divide the distance by the time to get the unit rate. (The amount they drove each hour.)

Maren: 60 / 1 = 60

Safiya: 120 / 3 = 40

Tomas: 84 / 1.2 = 70

Cam: 75 / 1.5 = 50

Tomas has the largest unit rate therefore he is  the fastest.

Solnce55 [7]3 years ago
5 0

Answer:

Tomas

Step-by-step explanation:

Divide Distance by Time

\frac{distance}{time} = <em>speed</em>

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The value of the truck initially, Ao is
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1-0.16=0.84
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After one year the value
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Hope it helps
8 0
3 years ago
Please help ASAP!!!
QveST [7]
The answer is 3. 1/6 because there are 6 sides in the cube. So the probability of getting a certain number is 1/6.
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What is the smallest value in the stem and leaf plot below?
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3 years ago
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
ASAP PLEASE HELP NO LINKS
ArbitrLikvidat [17]

Answer:

a is 3,2 b is 1,-2 and c is -2,1

Step-by-step explanation:

u take the first number then you just put em together

5 0
2 years ago
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