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3 years ago

Answer please thanks

Mathematics

1 answer:

Gnesinka [82]3 years ago

6 0

5 and a half that might be the answer

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Solve 3 over x minus 3 equals the quantity of x over x minus 3, minus three-halves. for x and determine if the solution is extra

astraxan [27]

$\frac{3}{x} -3 = \frac{x}{x} - \frac{3}{2}$
$3 -3x = x - \frac{3x}{2}$
$-6x -2x -x = 6$
$-9x=6$
$9x=6$
$x= \frac{6}{9}= \frac{2}{3}$

3 0

2 years ago

The table below shows the budget for the Smith family.

tiny-mole [99]

Answer:

415 300 213 3400 4500 = 8415

4500 450 415 300 215 3400 = 9280

8828 times 100 divided by 9280 =9 5.1293103448

95.1 percent (simplified)

Step-by-step explanation: income is income rescieved, id.k if income taxes is given to you but taxes is usually taken away from you, college savings, retirement saving and emergency funds are income, and living expenses is for living and you use it on what ever you want. mainly food.

8 0

3 years ago

Read 2 more answers

Trevor and marissa together have 26 t-shirts to sell. If marissa has 6 fewer tshirts that trevor, find hiw many tshirts trevor h

Eva8 [605]

Answer:

Number of t-shirts Trevor have are 16

If Trevor have 16 t-shirts, than Marissa will have 10 t-shirts.

Step-by-step explanation:

Let, the number of t-shirts Trevor has = x and number of t-shirts Marissa has = y.

Since, they both have total of 26 t-shirts. We get,

$x+y=26$

Moreover, Marissa has 6 t-shirts less than that of Trevor.

So, we have,

$y=x-6$

On solving the system of equations, we get,

$x+y=26$ implies $x+x-6=26$ implies $2x=26+6$ i.e. 2x= 32 i.e. x= 16.

So, $y=x-6$ implies $y=16-6$ i.e y= 10.

Hence, the number of t-shirts Trevor and Marissa have are 16 and 10 respectively.

That is, If Trevor have 16 t-shirts, than Marissa will have 10 t-shirts.

8 0

2 years ago

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iragen [17]

What is that? And what you need help for

6 0

2 years ago

Read 2 more answers

Find the absolute extreme of the function on the closed interval. y=9e^xsinx , [0,pi]

Molodets [167]

For convenience sake, I will let $y=f(x)=9e^{x}\sin x$

First, we evaluate the function at the endpoints of the interval.

$f(0)=9e^{0}\sin(0)=0\\\\f(\pi)=9e^{\pi}\sin(\pi)=0$

Then, we need to find the critical points.

We can start by taking the derivative using the power rule.

$f'(x)=9e^{x} \frac{d}{dx} \sin x+9\sin x \frac{d}{dx} e^{x}=9e^{x}(\cos x+\sin x)$

Setting this equal to 0,

$9e^x (\cos x+\sin x)=0$

Since $9e^x > 0$ , we can divide both sides by $9e^x$ .

$\cos x+\sin x=0\\\\\sin x=-\cos x\\\\\tan x=-1\\\\x=\frac{3\pi}{4}$

$f\left(\frac{3\pi}{4} \right)=9e^{3\pi/4}\sin \left(\frac{3\pi}{4} \right)=\frac{9\sqrt{2}e^{3\pi/4}}{2}$

So, the absolute minimum is $\boxed{\left(\frac{3\pi}{4}, \frac{9\sqrt{2}e^{3\pi/4}}{\sqrt{2}} \right)}$ and the absolute minima are $\boxed{(0,0), (\pi, 0)}$

3 0

1 year ago