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harina [27]
2 years ago
15

the graph shows the triangle abc and the triangle def what is the scale factor used to create triangle def

Mathematics
1 answer:
BlackZzzverrR [31]2 years ago
5 0
The scale factor is positive 4.
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Step-by-step explanation:

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Can someone explain to me what a “derivative” means? How do you find the derivative of f(x)=x^3+1?
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The derivative is the rate of change of a function, basically represents the slope at different points. To find the derivative of the given function you can use the power rule, which means, if n is a real number, d/dx(x^n)= nx^(n-1). This is a simplification of the chain rule based on the fact that d/dx(x)=1. Anyway, this means that d/dx(x^3 + 1)= 3x^2. Here n is 3 and so it is 3*x^(3-1)= 3x^2. The derivative of x^3+1 is 3x^2.

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Read 2 more answers
39-50 find the limit.<br> 41. <img src="https://tex.z-dn.net/?f=%5Clim%20_%7Bt%20%5Crightarrow%200%7D%20%5Cfrac%7B%5Ctan%206%20t
Katyanochek1 [597]

Write tan in terms of sin and cos.

\displaystyle \lim_{t\to0}\frac{\tan(6t)}{\sin(2t)} = \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)}

Recall that

\displaystyle \lim_{x\to0}\frac{\sin(x)}x = 1

Rewrite and expand the given limand as the product

\displaystyle \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)} = \lim_{t\to0} \frac{\sin(6t)}{6t} \times \frac{2t}{\sin(2t)} \times \frac{6t}{2t\cos(6t)} \\\\ = \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right)

Then using the known limit above, it follows that

\displaystyle \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right) = 1 \times 1 \times \frac3{\cos(0)} = \boxed{3}

4 0
2 years ago
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