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Harrizon [31]
4 years ago
12

Solve the following inequality:

Mathematics
1 answer:
ad-work [718]4 years ago
4 0

Answer:

The answer is option 3.

Step-by-step explanation:

In order to solve the inequality, you have to get rid of -3 by adding 3 to both sides :

k - 3 \leqslant 9

k - 3 + 3 \leqslant 9 + 3

k \leqslant 12

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4 years ago
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8 0
3 years ago
HELP ME PLS explain why
Anika [276]

Answer: 110 square cm

=======================================================

Explanation:

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3 0
3 years ago
−4x+7y+5=0<br> x−3y=−5<br> ​How many solutions does the system have?
oksano4ka [1.4K]
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4 0
3 years ago
(15 Points)
expeople1 [14]

ANSWER 1



Note that,


f(u)=tan^{-1}(u)


is the same as


f(u)=arctan(u)



We apply the product rule.


f(x)=x^2tan^{-1}(x)


So we keep the second function and differentiate the first,plus we keep the first function and differentiate the second.


f'(x)=(x^2)'tan^{-1}(x)+x^2(tan^{-1}(x))'



Recall that,

If

f(u)=tan^{-1}(u)



Then,

f'(u)=\frac{1}{1+u^2}} \times u'


This implies that,

f'(x)=2xtan^{-1}(x)+\frac{x^2}{x^2+1}



ANSWER 2


We apply the product rule and the chain rules of differentiation here.



f(x)=xsin^{-1}(1-x^2)




f'(x)=x'sin^{-1}(1-x^2)+x(sin^{-1}(1-x^2))'



Recall that,

If

f(u)=sin^{-1}(u)



Then,

f'(u)=\frac{1}{\sqrt{1-u^2}} \times u'



This implies that,


f'(x)=sin^{-1}(1-x^2)+x \times \frac{1}{\sqrt{1-(1-x^2)^2}}\times (-2x)


f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-(1-2x^2+x^4)}}


f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-1+2x^2-x^4}}



f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{2x^2-x^4}}





5 0
4 years ago
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