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1 year ago

Which polynomial represents twice the sum of -5x^2-x+6 and 7x^2+x-1?

Mathematics

1 answer:

elena-14-01-66 [18.8K]1 year ago

7 0

The polynomial that represent twice the sum of $-5x^{2}$ - x +6 and 7 $x^{2}$ + x -1 is : 4 $x^{2}$ +10

<h3>What is a polynomial?</h3>

A polynomial is any algebraic expression that has the least value of the power of its variable as 2.

When the highest value of the polynomial is 2, it is a called a quadratic expression.

Analysis:

2( -5 $x^{2}$ -x +6 + 7 $x^{2}$ +x -1)

2( 2 $x^{2}$ + 5) =

4 $x^{2}$ + 10

In conclusion, the polynomial that represents the above expressions is 4 $x^{2}$ + 10

Learn more about polynomials: brainly.com/question/2833285

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Determine if the following table represents a quadratic function.

snow_tiger [21]

Answer:

The options are not visible enough (See Explanation)

Step-by-step explanation:

Given

x:- 1 || 2 || 3 || 4 || 5

y:- 13 || 22 || 37 || 58 || 85

Required

Determine if the function is quadratic

Calculate the difference between the values of y

$Difference = 22 - 13 = 9$

$Difference = 37 - 22 = 15$

$Difference = 58 - 37 = 21$

$Difference = 85 - 58 = 27$

The resulting difference are: 9 || 15 || 21 || 27

Next; Calculate the difference between the difference of values of y

$Difference = 15 - 9 = 6$

$Difference = 21 - 15 = 6$

$Difference = 27 - 21 = 6$

The resulting difference are: 6 || 6 || 6

For the function to be quadratic, the above difference must be the same and since they are the same (6), then the function represents a quadratic function.

7 0

3 years ago

Find derivative problem Find B’(6)

dalvyx [7]

Answer:

$B^\prime(6) \approx -28.17$

Step-by-step explanation:

We have:

$\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)$

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

$\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]$

We can move the constant outside:

$\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]$

Now, we will utilize the product rule. The product rule is:

$(uv)^\prime=u^\prime v+u v^\prime$

We will let:

$\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t$

Then:

$\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1$

(The derivative of u was determined using the chain rule.)

Then it follows that:

$\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}$