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2 years ago

Which equation has the correct sign on the product?

Mathematics

2 answers:

Alla [95]2 years ago

4 0

Answer:

Step-by-step explanation:

A positive times a negative number equals a negative number, but a negative times a negative equals a positive number

expeople1 [14]2 years ago

3 0

C) 52 x 2 = 104

Remember:

Positive x positive = positive

Negative x negative = positive

Positive x negative = negative

Negative x positive = negative

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I need help with this

balandron [24]

Answer:

Step-by-step explanation:

Area of Triangle x 2 = 3 * 3

The reason I did not divide it by 2 is because of how the other triangle is bound to have the same area. Just leave it as it is because later you're going to reach the same answer.

3 * 3 = 9

Rectangle Area = 6 * 3 = 18

18 + 9 = 27

6 0

2 years ago

Mr. Campbell. paid $48 for 12 cartons of juice.

Wewaii [24]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

The right triangle on the right is a scaled

Whitepunk [10]

Answer:

The scale factor is 3.5

6 0

2 years ago

2c+17.6 =6 / 4 SOLVE give your answer as a decimal get brainly

tamaranim1 [39]

Answer:

-8.05

Step-by-step explanation:

In order to solve for C you first have to subtract 17.6 from both sides.

2c=(6/4)-17.6

6/4 is 1.5

1.5-17.6=-16.1

2c=-16.1

c=-8.05

8 0

2 years ago

Read 2 more answers

How to find the derivative of cos^2x? i seem to be confused.

slamgirl [31]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2927231

————————

You can actually use either the product rule or the chain rule for this one. Observe:

• Method I:

y = cos² x

y = cos x · cos x

Differentiate it by applying the product rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x\cdot cos\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x)\cdot cos\,x+cos\,x\cdot \dfrac{d}{dx}(cos\,x)}$

The derivative of cos x is – sin x. So you have

$\mathsf{\dfrac{dy}{dx}=(-sin\,x)\cdot cos\,x+cos\,x\cdot (-sin\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=-sin\,x\cdot cos\,x-cos\,x\cdot sin\,x}$

$\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark$

—————

• Method II:

You can also treat y as a composite function:

$\left\{\!
\begin{array}{l}
\mathsf{y=u^2}\\\\
\mathsf{u=cos\,x}
\end{array}
\right.$

and then, differentiate y by applying the chain rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}\\\\\\
\mathsf{\dfrac{dy}{dx}=\dfrac{d}{du}(u^2)\cdot \dfrac{d}{dx}(cos\,x)}$

For that first derivative with respect to u, just use the power rule, then you have

$\mathsf{\dfrac{dy}{dx}=2u^{2-1}\cdot \dfrac{d}{dx}(cos\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=2u\cdot (-sin\,x)\qquad\quad (but~~u=cos\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=2\,cos\,x\cdot (-sin\,x)}$

and then you get the same answer:

$\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark$

I hope this helps. =)

Tags: <em>derivative chain rule product rule composite function trigonometric trig squared cosine cos differential integral calculus</em>

3 0

3 years ago