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4 years ago

I don’t know how to answer this

Mathematics

1 answer:

masya89 [10]4 years ago

4 0

Answer:

20 ≥ 9 + (8 -7)·6
20 ≥ 1 + (2 -3)·4
20 ≥ 8 +(3 -5)·4

Step-by-step explanation:

You mainly need to make sure that the product has a value less than 20. That's not hard to do, since the difference can't be more than 9, and can be negative.

Pick some digits to put into the expression and see what the value is. Then make any adjustments you might need to make to reduce the sum. If you put 0 in the rightmost box, the other digits can be anything. (They just have to be different.)

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BlackZzzverrR [31]

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Use the table below to evaluate the derivative with respect to x of g of f of 2 times x at x = 1.

lilavasa [31]

(g o f)(2x)=g(f(2x))

take derivitive
use chain rule
remember
derivitive of g(f(x))=g'(f(x))f'(x)
so

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at x=1
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we get
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it is 2
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3 years ago

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Answer:

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Step-by-step explanation:

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FromTheMoon [43]

Answer:

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Step-by-step explanation:

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3 years ago

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How do you find the length of the curve x=3t−t3x=3t-t^3, y=3t2y=3t^2, where 0≤t≤3√0<=t<=sqrt(3) ?

taurus [48]

It's simple, you just have to do this:

$L=\int\limits_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}~dt$

$x=3t-t^3$

$\frac{dx}{dt}=3-3t^2$

$y=3t^2$

$\frac{dy}{dt}=6t$

replacing

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{\left(3-3t^2\right)^2+\left(6t\right)^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{9-18t^2+9t^4+36t^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{9+9t^4+18t^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{(3t^2+3)^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}(3t^2+3)~dt$

$L=t^3+3t|_0^{\sqrt{3}}$

$\boxed{\boxed{L=3\sqrt{3}+3\sqrt{3}=6\sqrt{3}}}$

8 0

3 years ago