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3 years ago

What is the answer for 47

Mathematics

1 answer:

bazaltina [42]3 years ago

7 0

A, because the equation shows that G is opposite number for P.

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Uma pessoa deseja restaurar seu veículo antigo e para isso deverá contratar serviços mecânico, de funilaria e de autoelétrica. P

Leni [432]

Answer:

CI=P*(1 + R/100)^18

A=(CI + P) = P(1+R/100)^18

13500/P=1(100+R/100)^18

A/P=(100+R/100)^18

A=13500$ as (750 * 18)

(13500)/P=(1 +1.15/100)18

(13500)/P=(1+1.15/100)18

13500=((1.0115)^18

P=R$10989.02

Step-by-step explanation:

CI=Compound Interest

A=Amount

P=Principal.

3 0

3 years ago

Zoe wrote 3/5 of her essay in 30 minutes. Assuming she wrote the paper at a constant rate the entire time, which expression can

LUCKY_DIMON [66]

Answer: The answer is (B)

Step-by-step explanation:

It just took the unit test

3 0

3 years ago

The distances traveled by car A car Afx x hours. are. represented by the graph and table below.<br>

jarptica [38.1K]

Answer: b

Step-by-step explanation:

3 0

2 years ago

14x - 6 - 7x + 10 = -17

Luden [163]

X equals -3. if you need the steps just search up the equation!

4 0

2 years ago

Read 2 more answers

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

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

7 0

3 years ago