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mrs_skeptik [129]

3 years ago

15

If the cost of a medium shirt is $10 and a small is $8, Make a table to find the cost for all

1 answer:

Annette [7]3 years ago

3 0

<h3>Answer: I can't draw on this though</h3>

Step-by-step explanation:

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Solve the initial value problem y'=2 cos 2x/(3+2y),y(0)=−1 and determine where the solution attains its maximum value.

zloy xaker [14]

Answer:

$y=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

$x={\pi}{4}$

Step-by-step explanation:

We are given that

$y'=\frac{2cos2x}{3+2y}$

y(0)=-1

$\frac{dy}{dx}=\frac{2cos2x}{3+2y}$

$(3+2y)dy=2cos2x dx$

Taking integration on both sides then we get

$\int (3+2y)dy=2\int cos 2xdx$

$3y+y^2=sin2x+C$

Using formula

$\int x^n=\frac{x^{n+1}}{n+1}+C$

$\int cosx dx=sinx$

Substitute x=0 and y=-1

$-3+1=sin0+C$

$-2=C$

$sin0=0$

Substitute the value of C

$y^2+3y=sin2x-2$

$y^2+3y-sin 2x+2=0$

$y=\frac{-3\pm\sqrt{(3)^2-4(1)(-sin2x+2)}}{2}$

By using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$y=\frac{-3\pm\sqrt{9+4sin2x-8}}{2}=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

Hence, the solution $y=\frac{-3\pm\sqrt{4sin2x+1}}{2}$

When the solution is maximum then y'=0

$\frac{2cos2x}{3+2y}=0$

$2cos2x=0$

$cos2x=0$

$cos2x=cos\frac{\pi}{2}$

$cos\frac{\pi}{2}=0$

$2x=\frac{\pi}{2}$

$x=\frac{\pi}{4}$

3 0

3 years ago

Monroe started saving money. Right now he

Ray Of Light [21]

Answer:

After 4 months

Step-by-step explanation:

200+15+15+15+15=260

300-10-10-10-10=260

6 0

2 years ago

Read 2 more answers

Marla thinks her dog is not feeling well. He ate 0.1 times As much dog food on Wednesday as he ate on Tuesday . If her dog ate 6

Anvisha [2.4K]

5.9 is how much he ate

4 0

2 years ago

Read 2 more answers

What is the solution to the inequality 2|5-2x|-3 less than or equal to 15?

vazorg [7]

× =< 11\2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

5 0

3 years ago

An object is propelled upward from the top of a 300 foot building. The path that the object takes as it falls to the ground can

8090 [49]

Answer:

As per the statement:

The path that the object takes as it falls to the ground can be modeled by

h =-16t^2 + 80t + 300

where

h is the height of the objects and

t is the time (in seconds)

At t = 0 , h = 300 ft

When the objects hit the ground, h = 0

then;

-16t^2+80t+300=0

For a quadratic equation: ax^2+bx+c=0 ......[1]

the solution for the equation is given by:

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

On comparing the given equation with [1] we have;

a = -16 ,b = 80 and c = 300

then;

$t= \frac{-80\pm \sqrt{(80)^2-4(-16)(300)}}{2(-16)}$

$t= \frac{-80\pm \sqrt{6400+19200}}{-32}$

$t= \frac{-80\pm \sqrt{25600}}{-32}$

Simplify:

$t = -\frac{5}{2}$ = -2.5 sec and $t = \frac{15}{2}$ = 7.5 sec

Time can't be in negative;

therefore, the time it took the object to hit the ground is 7.5 sec

6 0

3 years ago

Read 2 more answers