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2 years ago

The price of an item has been reduced by 65% . The original price was $49

2 answers:

6 0

65% x
100 49
Cross multiply And divide 100 because that's the only number
Left alone and 65 is your percent so you have to put it over 100
X is your variable, your unknown number multiply 65 times 49
divided by 100 because that's your only number ! Hope this helps

Basile [38]2 years ago

4 0

Hey there!

To find the discount, just multiply the original price by the percent.

49*.65= 31.85(how much it was taken off)

Now subtract31.85 from 49 and you'll get
17.15

$17.15 is now the new price

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Which of the following functions are solutions of the differential equation y'' + y = 3 sin x? (select all that apply)

DiKsa [7]

Answer:

Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. $y'=(3\sin x)'=3\cos x$ . Differentiate again to get $y''=(3\cos x)'=-3\sin x$ , then $y''+y=-3\sin x+3\sin x=0\neq 3\sin x$ so this choice of y doesn't solve the equation.

B. $y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x$ and $y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x$ , then $y''+y=10\sin x+6\cos x\neq 3\sin x$ so y is not a solution

C. $y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x$ hence $y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x$ . Then $y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x$ so y is a solution.

D. $y'=-3\sin x$ and $y''=-3\cos x$ , then $y''+y=0$ thus y isn't a solution

E. $y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x$ hence $y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x$ . Then $y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x$ then y is not a solution.

3 0

3 years ago

Is 20 quart greater than 80 cups

White raven [17]

I think it's equal
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3 0

2 years ago

If someone could please walk me through on how to answer this.

Elis [28]

To isolate c means to separate it completely on one side of the equals sign.

To isolate variables, you apply opposite operations.

In E = mc², m and c are being multiplied together. To separate them, you divide by the variable you want to get rid of. However, you must do this to both sides of the equation always. Whatever you do to one side of the equation you must do to the other side as well. This is so the equation remains true.

Since we want to isolate c, we'll start by dividing both sides by m.

E = mc²

E/m = mc²/m

E/m = c² -- The m's cancel as 1

Now we have c squared. The opposite of squaring something is taking its square root. Take the square root of each side.

E/m = c²

√(E/m) = √(c²)

√(E/m) = c -- Opposite operations cancel each other out

And you've isolated c!

Answer:

c = √(E/m)

3 0

3 years ago

A sandbox has an area of 10 2/3 square feet, and the length is 1/4 feet. What is the width of the sandbox?

AlekseyPX

Answer:

The width of the sandbox is $42\frac{2}{3} \ ft$ .

Step-by-step explanation:

Given,

Area of the sandbox = $10\frac{2}{3}=\frac{32}{3}\ ft^2$

Length = $\frac{1}{4}\ ft$

Solution,

Let the width of the sandbox be 'w'.

Since the sandbox is in shape of rectangle.

So we use the formula of area of rectangle.

$Area = Length\times Width$

On substituting the given values, we get;

$\frac{1}{4}\times w=\frac{32}{3}\\$

By cross multiplication method, we get;

$w=\frac{32\times4}{3\times 1} =\frac{128}{3} =42\frac{2}{3} \ ft$

Hence The width of the sandbox is $42\frac{2}{3} \ ft$ .

5 0

3 years ago

What is the solution to the system of equations? 3x - 6y = -12, x -2y =-8.

Xelga [282]

Answer:

Step-by-step explanation:

First, we need to write down the equations so we can have a vision of the equations.

3x - 6y = -12

x - 2y = -8

Now we find the simpler equation, and we make it to where one of the variables is on one side.

Let's use the bottom equation.

x = 2y - 8

Now we plug in the equation to the other equation.

3(2y - 8) - 6y = -12

6y - 24 - 6y = -12

-24 = -12

0 = 12

This is a false statement. 0 does not equal 12. That means that the equation has no solutions.

6 0

1 year ago