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Anarel [89]
3 years ago
8

What is the product of (3x-9) and 6x↑2-2x+5 Write your answer in standard form.

Mathematics
1 answer:
klio [65]3 years ago
3 0

Answer:

a. 18x^3 -60x^2 + 33x - 45

b. No

Step-by-step explanation:

Given:

(3x-9) and 6x^2-2x+5

=> the production of them is:

(3x-9)(6x^2-2x+5) , we will use distributive law to solve this To do that. we first multiply  6x^2-2x+5 by 3x then by -9. This is done as follows

= 3x (6x^2-2x+5) - 9(6x^2-2x+5)  = (3x * 6x^2-3x * 2x+3x *5) - (9* 6x^2-9* 2x+9* 5)\\\\= (18x^3 -6x^2+ 15x) - (54x^2 - 18x +45)

Then Open both brackets

= 18x^3 -6x^2+ 15x -54x^2 + 18x -45

After that, we group the same terms

= 18x^3 -(6x^2 +54x^2)+ (15x  + 18x) -45

= 18x^3 -60x^2 + 33x - 45

(b) Is the product of (3x-9) and  6x^2-2x+5  equal to the product of (9x-3) and 6x^2-2x+5

No, let find the product of (9x-3) and 6x^2-2x+5 , we will use distributive law to solve this as the above example.

= (9x-3)(6x^2-2x+5)

= 9x (6x^2-2x+5) - 3(6x^2-2x+5)

=(9x * 6x^2 - 9x * 2x + 9x *5) - (3* 6x^2 - 3 * 2x + 3* 5)

= (54x^3  - 18x^2+ 45x) - (18x^2 - 6x +15)

After that, we open the bracket to find the same terms

= 54x^3  - 18x^2+ 45x -18x^2 + 6x -15

= 54x^3  - 18x^2 -18x^2 + 45x  + 6x -15

= 54x^3  - 36x^2 + 51x  -15  

As you can see 54x^3  - 36x^2 + 51x  -15   ≠18x^3 -60x^2 + 33x - 45

Hope it will find you well.

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2 years ago
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Kisachek [45]

Answer:

\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2}  t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2

Step-by-step explanation:

\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2}  t^2+1 \ \text{dt} = \ ?

We can use Part I of the Fundamental Theorem of Calculus:

  • \displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

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We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

  • \displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

  • \displaystyle\int\limits^b_a \text{f(t) dt}\  = -\int\limits^a_b \text{f(t) dt}

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

  • \displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx}  \int\limits^{x^2}_0 t^2+1 \text{ dt}  

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

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For the first term, replace \text{t} with 2x, and apply the chain rule to the function. Do the same for the second term; replace

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  • -8x^2 -2 + 2x^5 + 2x

Rearrange the terms to be in order from the highest degree to the lowest degree.

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This is the derivative of the given integral, and thus the solution to the problem.

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