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3 years ago

Daniel calculated that the circumference of his motorcycle tire is 62.8 inches. What is the tire's radius? Use 3.14 for pi.

Mathematics

1 answer:

Savatey [412]3 years ago

6 0

R=19.98 In by dividing

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If an angle is acute, then the measure of its complement must be greater than the measure of its

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Answer: acute angles are higher that 0° and smaller than 90°

Step-by-step explanation: obtuse are between 90° and 180°

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2 years ago

Curtis paid $80 on the vet bill for his dog. Then the vet charged him $40 more for his dog's medicine. If he now owes the vet $1

Klio2033 [76]

Answer:

He paid 80 at the start, so his starting must have been 80

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80+40=120...

so either 0, or 80 because he owed nothing, or he owed the 80, then the 40 for medication

I hope this helps!

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2 years ago

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Which equation is false?

aivan3 [116]

C is false. 9 multiplied by 7 equals 63.

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2 years ago

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Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

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3 years ago

Need help on 8,9,10 I don’t know it and my work is due tmr

FrozenT [24]

Answer:

36 right

Step-by-step explanation:

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2 years ago

Daniel calculated that the circumference of his motorcycle tire is 62.8 inches. What is the tire's radius? Use 3.14 for ​pi.

Daniel calculated that the circumference of his motorcycle tire is 62.8 inches. What is the tire's radius? Use 3.14 for pi.