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3 years ago

Solve for x. 2cosxsinx=0

1 answer:

8 0

The first step would be to divde both sides by 2. then you would have to find where cos(x) times sin(x) would equal 0. to do that either cos(x) OR sin(x) would heve to be zero becasue any number times zero equals zero. n is an inteager (whole Number) cos(x) equals pi/2 plus pi times n. $2cos(x)sin(x)=0 \\ cos(x)sin(x)=0 \\ cos(x)=0 \\( \pi /2) + \pi n\\ sin(x)=0 \\ 0 + \pi n$

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ASAP HELP you have $125 dollars in a savings account and you deposit an equal amount into your account each week. After 5 weeks

Ronch [10]

Answer:

Step-by-step explanation:

7 0

3 years ago

Jack and Diane were saving money to go to a rock n roll concert. They need at least $750 in order to go if you count their gas,

nataly862011 [7]

Answer:

$\left\{\begin{array}{l}y\le 4x\\ \\x\ge 10\\ \\50x+25y\ge 750\end{array}\right.$

Step-by-step explanation:

Let x represents the number of nights Jack worked and y represents the number of nights Diane worked.

1. The number of nights Diane is scheduled to work is no more than four times the number of nights Jack is scheduled to play. Then

$y\le 4x$

2. Diane will work at least 10 times before the concert. Then

$x\ge 10$

3. Jack earns $50 per night that he plays, then he earned $50x in x nights. Diane earns $25 each night she works, then she earns $25y in y nigths. They need at least $750, so

$50x+25y\ge 750$

4. We get the following set of constraints to model the problem:

$\left\{\begin{array}{l}y\le 4x\\ \\x\ge 10\\ \\50x+25y\ge 750\end{array}\right.$

4 0

3 years ago

Find the absolute maximum and minimum values of f(x,y)=xy−4x in the region bounded by the x-axis and the parabola y=16−x2.

skad [1K]

Answer:

The absolute maximum and minimum is $20\; \text{and} -20.$

Step-by-step explanation:

We first check the critical points on the interior of the domain using the

first derivative test.

$f_x=y-4=0$

$f_y=x=0$

The only solution to this system of equations is the point (0, 4), which lies in the domain.

$f_{xx}=0, \;f_{yy}=0\; \text{and}\; f_{xy}=-1$

$\Rightarrow f_{xx}f_{yy}-f_{xy}=o-1=-1$

$\therefore (0,4)$ is a saddle point.

Boundary points - $(4,0), (-4,0), (0,16)$

Along boundary $y=16-x^2$

$f=x(16-x^2)-4x$

$=16x-x^3-4x$

$\Rightarrow f^'=16-3x^2-4=0$

$\Rightarrow 3x^2=12$

$\Rightarrow x=\pm2,\;\;y=14$

Values of f(x) at these points.

$\begin{array}{}(4,0)=-16\\(-4,0)=16\\(0,16)=0\\(2,14)=20\\(-2,14)=-20\end{array}$

Therefore, the absolute maximum and minimum is $20\; \text{and} -20.$

6 0

3 years ago

What is the slope? (5,6) and (-1, 4)

7nadin3 [17]

Answer:

1/3

Step-by-step explanation:

y2-y1/x2-x1

4-6/-1-5

-2/-6

1/3

3 0

3 years ago

Read 2 more answers

Identify the diameter of the circular base created by folding the figure into a right cone. HELP ASAP PLEASE!!

Akimi4 [234]

let's notice something, we have a circle with a radius of 12 and one 90° sector is cut off, so only three 90° sectors of the circle are left shaded, so namely the cone will be using 3/4 of that circle.

think of it as, this shaded area is some piece of paper, and you need to pull it upwards and have the cutoff edges meet, and when that happens, you'll end up with a cone-shaped paper cup, and pour in some punch.

now, once we have pulled up the center of the circle to make our paper cup, there will be a circular base, its diameter not going to be 24, it'll be less, but whatever that base is, we know that is going to have the same circumference as those in the shaded area. Well, what is the circumference of that shaded area?

$\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=12 \end{cases}\implies C=2\pi 12\implies C=24\pi \implies \stackrel{\textit{three quarters of it}}{24\pi \cdot \cfrac{3}{4}} \\\\\\ 6\pi \cdot 3\implies 18\pi$

well then, the circumference of that circle at the bottom will be 18π, so, what is the diameter of a circle with a circumferenc of 18π?

$\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=18\pi \end{cases}\implies 18\pi =2\pi r\implies \cfrac{18\pi }{2\pi }=r\implies 9=r \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{diameter is twice the radius}}{d=18}~\hfill$

3 0

3 years ago