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Alexandra [31]
3 years ago
13

PLZ ANSWER !!! QUICKLY

Mathematics
1 answer:
a_sh-v [17]3 years ago
4 0
The answer is 560in^3
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What are the critical points for the inequality x^2 - 4/x^2 - 5x +6 < 0?
Leokris [45]

Answer:

Critical points are: 2 , -2 and 3.

Step-by-step explanation:

We have been given the expression:

\frac{x^2-4}{x^2-5x+6}

We will factorize the given expression:

Using a^2-b^2=(a+b)(a-b)

\frac{(x+2)(x-2)}{x^2-3x-2x+6}

\Rightarrow \frac{(x+2)(x-2)}{x(x-3)-2(x-3)}

\Rightarrow \frac{(x+2)(x-2)}{(x-2)(x+3)}

Critical point are those values of an expression when it is equal to zero

Hence, the critical points are: 2,-2 and -3.


7 0
4 years ago
Read 2 more answers
#15 & #16 please thanks
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Answer:

15. 122.49  16.13.17

Step-by-step explanation:

6 0
4 years ago
1 + 10 square theta (1 - sin square theta is equals to 1<br>​
KengaRu [80]

Let's prove

\boxed{\sf 1+tan^2\theta(1-sin^2\theta)=1}

LHS

\\ \sf\longmapsto 1+tan^2\theta(1-sin^2\theta)

\\ \sf\longmapsto 1+\dfrac{sin^2\theta}{cos^2\theta}(1-sin^2\theta)

\\ \sf\longmapsto \dfrac{cos^2\theta+sin^2\theta}{cos^2\theta}(1-sin^2\theta)

\\ \sf\longmapsto \dfrac{1}{cos^2\theta}(1-sin^2\theta)

\\ \sf\longmapsto \dfrac{1-sin^2\theta}{cos^2\theta}

\\ \sf\longmapsto \dfrac{1-sin^2\theta}{1-sin^2\theta}

\\ \sf\longmapsto 1

Hence provee

3 0
3 years ago
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