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4 years ago

(3x2-10x+4)+(10x-5x+8)

Mathematics

1 answer:

Luba_88 [7]4 years ago

8 0

Answer:

13x^2 - 15x + 12

Step-by-step explanation:

Add like terms.

3x^2 + 10x^2 = 13x^2

-10x + -5x = -15x

4 + 8 = 12

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From a train station, one train heads north, and another heads east. Some time later, the northbound train has traveled 27 kilom

galben [10]

Answer: the eastbound train had travelled 12 miles
Explanation: using Pythagoras theorem, we know that
X^2 = a^2 + b^2
Where ^ stands for raised to power.
Let
a stand for the train going towards North and
b stands for the train going towards east.
X stands for the total distance between train a and b = 20 miles.
By Pythagoras rule
X^2 = a^2 + b^2
20^2 = 16^2 + b^2
400 = 256 + b^2 so that
b^2 = 400 - 256 = 144
b = √144
b = 12 miles.

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3 years ago

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Mumz [18]

Answer:

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4 years ago

For what values of a and b is the line 3x y=b tangent to the curve y=ax3 when x=3?

Sav [38]

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3 years ago

A video company randomly selected 100 of its subscribers and asked them how many hours of shows they watch per week. Of those su

Alik [6]

Answer:

1,135 people

For every 100 people only 45 people watch more than 10 hours per week. So, the ratio is (45/100), since the company has 2500 we want to know what (x/2500) is. The total ratio becomes (1,135/2,500).

The number is found by dividing 2500 by 100 to receive the common factor of 25. You multiply 45 by 25 to receive your total answer.

Step-by-step explanation:

For every 100 people only 45 people watch more than 10 hours per week. So, the ratio is (45/100), since the company has 2500 we want to know what (x/2500) is. The total ratio becomes (1,135/2,500).

The number is found by dividing 2500 by 100 to receive the common factor of 25. You multiply 45 by 25 to receive your total answer.

8 0

3 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

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

5 0

2 years ago