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3 years ago

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 depreciating value of a semi-truck can be modeled by y = Ao(0.84)x, where y is the remaining value of the semi, x is the tim

NemiM [27]

The value of the truck initially, Ao is
83000

1-0.16=0.84
1-0.26=0.74

After one year the value
Y=83,000×(0.84)=69,720
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When you compare the results you will see that the graph would fall at a faster rate to the right because the depreciation rate of 26% is higher than the depreciation rate of 16%

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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3 years ago

What is the smallest value in the stem and leaf plot below?

Alexeev081 [22]

Listing the data from the Stem and Leaves table as raw data we have:

02, 03, 03, 07, 10, 10, 17, 34, 34, 35, 35

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7 0

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