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1 year ago

A washer and a dryer cost $737 combined. The washer costs $37 more than the dryer. What is the cost of the dryer

Mathematics

1 answer:

jolli1 [7]1 year ago

7 0

Answer: $350

Step-by-step explanation:

$700 divided by 2 is 350

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Triangle Congruence Theorems

elixir [45]

Triangle Congruence Theorems

Use the triangle congruence theorems below to prove that two triangles are congruent if:
Three sides of one triangle are congruent to three sides of another triangle (SSS: side side side)
Two sides and the angle in between are congruent to the corresponding parts of another triangle (SAS: side angle side)
Two angles and the side in between are congruent to the corresponding parts of another triangle (ASA: angle side angle)
Two angles and the non-included side of one triangle are congruent to the corresponding parts of another triangle (AAS: angle angle side)
The hypotenuse and leg of one right triangle are congruent to the corresponding parts of another right triangle (HL: hypotenuse leg)

3 0

2 years ago

Estimate the sum: 1 3/7 + 3 5/9 + 1/2. Show your work.

NISA [10]

Add 1 3/7 and leave it like that, and then do 9x3=27+5=32 so it’s gonna be 32/9 and then calculate the sum into 691/126 or u can divided more into 5 61/126

8 0

2 years ago

Read 2 more answers

Mr. Turner employer pays 100% of his medical insurance. He has rent of $800 and $600 for car payment, $100 light bill, and $300

docker41 [41]

Given :

Mr. Turner employer pays 100% of his medical insurance. He has rent of $800 and $600 for car payment, $100 light bill, and $300 car insurance. His payroll and income taxes are estimated to be $287.

To Find :

What is his necessary monthly salary to have a gross income to meet his total expenses including the payroll and income taxes.

Solution :

Minimum salary to bear all expenses is :

S = Basic expenses + Payroll and taxes

S = $( 800 + 600 + 100 + 300 + 287)

S = $2087.00

Therefore, monthly salary is $2087.00 .

Hence, this is the required solution.

8 0

2 years ago

How far did the cargo ship travel in one hour?

Grace [21]

Answer:

needs a image not much information working with

4 0

1 year ago

Help with q25 please. Thanks.

Westkost [7]

First, I'll make f(x) = sin(px) + cos(px) because this expression shows up quite a lot, and such a substitution makes life a bit easier for us.

Let's apply the first derivative of this f(x) function.

$f(x) = \sin(px)+\cos(px)\\\\f'(x) = \frac{d}{dx}[f(x)]\\\\f'(x) = \frac{d}{dx}[\sin(px)+\cos(px)]\\\\f'(x) = \frac{d}{dx}[\sin(px)]+\frac{d}{dx}[\cos(px)]\\\\f'(x) = p\cos(px)-p\sin(px)\\\\ f'(x) = p(\cos(px)-\sin(px))\\\\$

Now apply the derivative to that to get the second derivative

$f''(x) = \frac{d}{dx}[f'(x)]\\\\f''(x) = \frac{d}{dx}[p(\cos(px)-\sin(px))]\\\\ f''(x) = p*\left(\frac{d}{dx}[\cos(px)]-\frac{d}{dx}[\sin(px)]\right)\\\\ f''(x) = p*\left(-p\sin(px)-p\cos(px)\right)\\\\ f''(x) = -p^2*\left(\sin(px)+\cos(px)\right)\\\\ f''(x) = -p^2*f(x)\\\\$

We can see that f '' (x) is just a scalar multiple of f(x). That multiple of course being -p^2.

Keep in mind that we haven't actually found dy/dx yet, or its second derivative counterpart either.

-----------------------------------

Let's compute dy/dx. We'll use f(x) as defined earlier.

$y = \ln\left(\sin(px)+\cos(px)\right)\\\\y = \ln\left(f(x)\right)\\\\\frac{dy}{dx} = \frac{d}{dx}\left[y\right]\\\\\frac{dy}{dx} = \frac{d}{dx}\left[\ln\left(f(x)\right)\right]\\\\\frac{dy}{dx} = \frac{1}{f(x)}*\frac{d}{dx}\left[f(x)\right]\\\\\frac{dy}{dx} = \frac{f'(x)}{f(x)}\\\\$

Use the chain rule here.

There's no need to plug in the expressions f(x) or f ' (x) as you'll see in the last section below.

Now use the quotient rule to find the second derivative of y

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]\\\\\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{f'(x)}{f(x)}\right]\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-f'(x)*f'(x)}{(f(x))^2}\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2}\\\\$

If you need a refresher on the quotient rule, then

$\frac{d}{dx}\left[\frac{P}{Q}\right] = \frac{P'*Q - P*Q'}{Q^2}\\\\$

where P and Q are functions of x.

-----------------------------------

This then means

$\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} + \left(\frac{f'(x)}{f(x)}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} +\frac{(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2+(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\$

Note the cancellation of -(f ' (x))^2 with (f ' (x))^2

------------------------------------

Let's then replace f '' (x) with -p^2*f(x)

This allows us to form ( f(x) )^2 in the numerator to cancel out with the denominator.

$\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*f(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*(f(x))^2}{(f(x))^2} + p^2\\\\-p^2 + p^2\\\\0\\\\$

So this concludes the proof that $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2 = 0\\\\$ when $y = \ln\left(\sin(px)+\cos(px)\right)\\\\$

Side note: This is an example of showing that the given y function is a solution to the given second order linear differential equation.

7 0

2 years ago