I need helppp pleaseeeeee!!!!!!!!!!!!!!

Negative three over four times five over seven divided by negative one over two

LekaFEV [45]

Answer:

$1 \frac{1}{14}$

Step-by-step explanation:

$- \frac{3}{4} \times \frac{5}{7} \div - \frac{1}{2} = - \frac{15}{28} \times -2 = \frac{30}{28} = 1 \frac{2}{28} = 1 \frac{1}{14}$

I am joyous to assist you anytime.

5 0

3 years ago

Help quickly not much time

LenKa [72]

Answer:

The answer is x=2/3

6 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

Please answer and show how you did it!

Anna35 [415]

This is a parabola, first, locate the line of symmetry.
the line of symmetry is x=-b/2a
in this case, b=-2, a=-1, so the line of symmetry is x=-1
when x=-1, f(x)=-(-1)²-2(-1)-3=-2
locate the point (-1,-2) on the grid. this point is the vertex.
get two pairs of points with x=-1 as the symmetry line:
(0, -3) and (-2, -3); (1,-6) and (-3,-6)
connect these five points into a parabola, stick out at the ends because it will extend forever downward.

3 0

2 years ago

Can somebody help me with this quickly whoever answers first gets a thanks

Mariana [72]

G. From the origin move right 3 units,then up 6 units

H. From the origin move left 2 units,then up 11 units

J. From the origin move right 8 units, then down 10 units

K. From the origin move left 16 units, then down 20 units

L. Plot the first point at the origin then move up 5 units

M. From the origin move left 14 units and plot the next point at the origin

5 0

2 years ago