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2 years ago

Which polynomial has the highest degree?

Mathematics

1 answer:

lina2011 [118]2 years ago

7 0

Answer:

A 3x^12 - 2x^8 + x^5

Step-by-step explanation:

Degree of a polynomial:

The degree of a polynomial is given by the highest exponent of x.

In this question:

Option A: 12th degree

Option B: 1st degree

Option C: 3rd degree

Option D: 3rd degree

So the answer is given by option A.

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B is the midpoint of AC. C is the midpoint of AD. Find the coordinates of B and D given the coordinates of A (-2,5) and the coor

GalinKa [24]

B= (3,6) and D= (18,9)

Step-by-step explanation:

To find the midpoint of two given points add X1 and X2 together, then divide by 2 to get the X value of the midpoint and do the same thing for the Y value. That will get you B. From there you can see to get from point A to point C you add 10 to the x value and add 2 to the y value. Because C is the midpoint just double what you had to add to get point D (add 20 to x and add 4 to y)

4 0

2 years ago

Find derivative problem<br> 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:

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

Therefore:

$\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]$

By simplification:

$\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17$

So, the slope of the tangent line to the point (6, B(6)) is -28.17.

5 0

2 years ago

2. An expression is shown.

laiz [17]

Adding parentheses in the component $9\div 3$ of the expression may bring an output of 48.

<h3>Procedure - Application of hierarchy rules in a arithmetic expression</h3>

In this question we should make use of hierarchy rules represented by the use of parentheses. The parentheses oblige to make operations inside it before making it in the rest of the formula.

Now we decide to add parenthesis in the component $9\div 3$ such that the result of the entire expression is 48. We proceed to present the proof:

$2 \times 8 \times (9\div 3)$