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3 years ago

Select the factors of x2 − 10x + 25.

Mathematics

2 answers:

CaHeK987 [17]3 years ago

6 0

Answer:

(x-5)2

Step-by-step explanation:

you must solve the expression (x-5)2 as the form (a-b)2 = a2 -2ab + b2

So it becomes x2-10x+25

Elena-2011 [213]3 years ago

6 0

Answer: (x-5)(x-5) or (x-5)²

Step-by-step explanation:

(a-b)(a-b)= a² - 2ab + b²

→ x² - 10x - 25

→ √25 = 5

so a= x, b= 5

this gives us (x-5)(x-5)

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frosja888 [35]

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3 years ago

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3 0

3 years ago

Read 2 more answers

Solve the following inequality.<br> -4(1 - 5a) > -124

NemiM [27]

Answer:

a>−6

Step-by-step explanation:

−4(1−5a)>−124

Step 1: Simplify both sides of the inequality.

20a−4>−124

Step 2: Add 4 to both sides.

20a−4+4>−124+4

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8 0

2 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

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2 years ago

What is the ratio for each one?

blagie [28]

This is complicated because I’m typing on a phone, but
24:30 simplified is 4:5
30:54 simplified is 5:9

10:5 simplified is 2:1
5:15 simplified is 1:3

32:72 simplified is 4:9
72:104 simplified is 9:13

56:7 simplified is 8:1
7:63 simplified is 1:9

8 0

3 years ago