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3 years ago

Factor by grouping 6v^3-14v^2+15v-35

Mathematics

1 answer:

spayn [35]3 years ago

6 0

Answer:

(3v-7)(2v^2+5)

Step-by-step explanation:

To factor 6v^3-14v^2+15v-35 by grouping we are going to try pair to up the pair two terms and also the last two terms. Like this:

(6v^3-14v^2)+(15v-35)

Now from each we factor what we can:

2v^2(3v-7)+5(3v-7)

Now there are two terms: 2v^2(3v-7) and 5(3v-7).

These terms contain a common factor and it is (3v-7).

We are going to factor (3v-7) out like so:

2v^2(3v-7)+5(3v-7)

(3v-7)(2v^2+5)

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Hitman42 [59]

Answer:

8.333

Step-by-step explanation:

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3 years ago

Everbank Field, home of the Jacksonville Jaguars, is capable of seating 76,867 fans. The revenue for a particular game can be mo

Aleksandr [31]

Given:
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3 years ago

Simplify the expression shown. [(25 × 16) + (75 × 16)] ÷ (3 + (-3) – 10)

mojhsa [17]

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2 years ago

Read 2 more answers

There are 1,500 students in a school. 65% of them are girls. How many girls are there in the school?

Veronika [31]

First you have to multiply 1,500 by 65%. A way to do that is 1,500 x 65 and then divide by 100.
So 1,500 x 65 is 97,500.
97,500 divided by 100 is 975.
The final answer is that there are 975 girls in the school.

3 0

3 years ago

Help with 30 please. thanks.

Svet_ta [14]

Answer:

See Below.

Step-by-step explanation:

We have the equation:

$\displaystyle y = \left(3e^{2x}-4x+1\right)^{{}^1\! / \! {}_2}$

And we want to show that:

$\displaystyle y \frac{d^2y }{dx^2} + \left(\frac{dy}{dx}\right) ^2 = 6e^{2x}$

Instead of differentiating directly, we can first square both sides:

$\displaystyle y^2 = 3e^{2x} -4x + 1$

We can find the first derivative through implicit differentiation:

$\displaystyle 2y \frac{dy}{dx} = 6e^{2x} -4$

Hence:

$\displaystyle \frac{dy}{dx} = \frac{3e^{2x} -2}{y}$

And we can find the second derivative by using the quotient rule:

$\displaystyle \begin{aligned}\frac{d^2y}{dx^2} & = \frac{(3e^{2x}-2)'(y)-(3e^{2x}-2)(y)'}{(y)^2}\\ \\ &= \frac{6ye^{2x}-\left(3e^{2x}-2\right)\left(\dfrac{dy}{dx}\right)}{y^2} \\ \\ &=\frac{6ye^{2x} -\left(3e^{2x} -2\right)\left(\dfrac{3e^{2x}-2}{y}\right)}{y^2}\\ \\ &=\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\end{aligned}$