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Anon25 [30]
3 years ago
7

(Factorize): x²-5x + 6​

Mathematics
1 answer:
faltersainse [42]3 years ago
5 0
Factor x
2−5x+6x

using the AC method.
(x−3)(x−2)
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Plz with steps .. it's very hard can anyone plz
liubo4ka [24]

Answer:

Step-by-step explanation:

\displaystyle\  \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\frac{0}{0} \\\\we\ can \ use\ Hospital's\ Rule\\\\\\f(x)=\sqrt{2x}-\sqrt{3x-a}  \qquad  f'(x)=\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} \\\\g(x)=\sqrt{x} -\sqrt{a}  \qquad g'(x)=\dfrac{1}{2\sqrt{x}} \\\\\\\displaystyle\  \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\lim_{n \to a} \dfrac{\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}}  }{\dfrac{1}{2\sqrt{x}} }\\\\

\displaystyle \lim_{n \to a} \dfrac{2\sqrt{x} }{\sqrt{2x}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}  =\lim_{n \to a} \dfrac{2 }{\sqrt{2}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3*\sqrt{a} }{\sqrt{2a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3}{\sqrt{2}}\\\\\\=-\ \dfrac{1}{\sqrt{2}}\\\\

7 0
3 years ago
Use the correct order of operations to calculate5 1/2 * 4 1/2 -7/2/ 2 1/3​
IgorC [24]
If there are multiple operations at the same level on the order of operations off from left to right and you work like this first noticed that there are no parentheses or exponents so we moved to multiplication and division with any sense of parentheses
5 0
3 years ago
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What is the answer to 1.3 (6.3r - 4.2) = 66.9
liubo4ka [24]
R = 804/91


8.835

this is ur answer i how i helped
5 0
3 years ago
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<img src="https://tex.z-dn.net/?f=%20%28%7B%20%7Bx%7D%5E%7B2%7D%20%20-%204%7D%29%5E%7B5%7D%20%28%20%7B4x%20-%205%7D%29%5E%7B4%7D
Makovka662 [10]

Let u=x^2-4 and v=4x-5. By the product rule,

\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=\dfrac{\mathrm d(u^5)}{\mathrm dx}v^4+u^5\dfrac{\mathrm d(v^4)}{\mathrm dx}

By the power rule, we have (u^5)'=5u^4 and (v^4)'=4v^3, but u,v are functions of x, so we also need to apply the chain rule:

\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}

\dfrac{\mathrm d(v^4)}{\mathrm dx}=4v^3\dfrac{\mathrm dv}{\mathrm dx}

and we have

\dfrac{\mathrm du}{\mathrm dx}=2x

\dfrac{\mathrm dv}{\mathrm dx}=4

So we end up with

\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3

Replace u,v to get everything in terms of x:

\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3

We can simplify this by factoring:

10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3=2(x^2-4)^4(4x-5)^3\bigg(5x(4x-5)+8(x^2-4)\bigg)

=2(x^2-4)^4(4x-5)^3(28x^2-57)

7 0
3 years ago
Simplify (4x − 6) + (3x + 6). (2 points) a 7x b 7x − 12 c x d 7x + 12
aleksley [76]

4x-6+3x+6=

7x

THE ANSWER IS "A"

8 0
3 years ago
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