Please help ASAP I’m gonna fail

The figure is rotated 360° clockwise with the origin as the center of rotation. Which graph represents the rotated figure?

Scilla [17]

Given

The figure is rotated 360° clockwise with the origin as the center of rotation. Which graph represents the rotated figure?

Answer

After the rotation of 360 degrees, a figure comes back to original position

Option A is correct

4 0

1 year ago

Part A: Create a third-degree polynomial in standard form. How do you know it is in standard form? (5 points)

Svetlanka [38]

Answer:

(See explanation for further details)

Step-by-step explanation:

a) Let consider the polynomial $p(x) = 5\cdot x^{3} +2\cdot x^{2} - 6 \cdot x +17$ . The polynomial is in standards when has the form $p(x) = \Sigma \limit_{i=0}^{n} \,a_{i}\cdot x^{i}$ , where $n$ is the order of the polynomial. The example has the following information:

$n = 3$ , $a_{0} = 17$ , $a_{1} = -6$ , $a_{2} = 2$ , $a_{3} = 5$ .

b) The closure property means that polynomials must be closed with respect to addition and multiplication, which is demonstrated hereafter:

Closure with respect to addition:

Let consider polynomials $p_{1}$ and $p_{2}$ such that:

$p_{1} = \Sigma \limits_{i=0}^{m} \,a_{i}\cdot x^{i}$ and $p_{2} = \Sigma \limits_{i=0}^{n}\,b_{i}\cdot x^{i}$ , where $m \geq n$

$p_{1}+p_{2} = \Sigma \limits_{i=0}^{n}\,(a_{i}+b_{i})\cdot x^{i} + \Sigma_{i=n+1}^{m}\,a_{i} \cdot x^{i}$

Hence, polynomials are closed with respect to addition.

Closure with respect to multiplication:

Let be $p_{1}$ a polynomial such that:

$p_{1} = \Sigma \limits_{i=0}^{m} \,a_{i}\cdot x^{i}$

And $\alpha$ an scalar. If the polynomial is multiplied by the scalar number, then:

$\alpha \cdot p_{1} = \alpha \cdot \Sigma \limits_{i = 0}^{m}\,a_{i}\cdot x^{i}$

Lastly, the following expression is constructed by distributive property:

$\alpha \cdot p_{1} = \Sigma \limits_{i=0}^{m}\,(\alpha\cdot a_{i})\cdot x^{i}$

Hence, polynomials are closed with respect to multiplication.

4 0

3 years ago

12. What's the simplified form of x − 1 + 4 + 7x − 3?

Leviafan [203]

The first one is B and the second is D

3 0

4 years ago

Read 2 more answers

») Noah finds 14 caterpillars and puts them in 2 cages.

Allisa [31]

Answer:

7 Caterpillars

Step-by-step explanation:

14/2

7 0

3 years ago

CALCULUS - Find the values of in the interval (0,2pi) where the tangent line to the graph of y = sinxcosx is

Rufina [12.5K]

Answer:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

Step-by-step explanation:

We want to find the values between the interval (0, 2π) where the tangent line to the graph of y=sin(x)cos(x) is horizontal.

Since the tangent line is horizontal, this means that our derivative at those points are 0.

So, first, let's find the derivative of our function.

$y=\sin(x)\cos(x)$

Take the derivative of both sides with respect to x:

$\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)]$

We need to use the product rule:

$(uv)'=u'v+uv'$

So, differentiate:

$y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)]$

Evaluate:

$y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))$

Simplify:

$y'=\cos^2(x)-\sin^2(x)$

Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':

$0=\cos^2(x)-\sin^2(x)$

Now, let's solve for x. First, we can use the difference of two squares to obtain:

$0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))$

Zero Product Property:

$0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)$

Solve for each case.

Case 1:

$0=\cos(x)-\sin(x)$

Add sin(x) to both sides:

$\cos(x)=\sin(x)$

To solve this, we can use the unit circle.

Recall at what points cosine equals sine.

This only happens twice: at π/4 (45°) and at 5π/4 (225°).

At both of these points, both cosine and sine equals √2/2 and -√2/2.

And between the intervals 0 and 2π, these are the only two times that happens.

Case II:

We have:

$0=\cos(x)+\sin(x)$

Subtract sine from both sides:

$\cos(x)=-\sin(x)$

Again, we can use the unit circle. Recall when cosine is the opposite of sine.

Like the previous one, this also happens at the 45°. However, this times, it happens at 3π/4 and 7π/4.

At 3π/4, cosine is -√2/2, and sine is √2/2. If we divide by a negative, we will see that cos(x)=-sin(x).

At 7π/4, cosine is √2/2, and sine is -√2/2, thus making our equation true.

Therefore, our solution set is:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

And we're done!

Edit: Small Mistake :)

5 0

3 years ago