Hi Can you answer these questions Either you answer all or just one it’s ok Thx

Mathematics

1 answer:

MA_775_DIABLO [31]3 years ago

4 0

Answer:

I'm sorry if you can't read it. I hope I could still help.

Step-by-step explanation:

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A box of crayons costs $1.75, including tax. Mr. Valentino wants to purchase boxes of crayons for his class and has a $25 budget

Mnenie [13.5K]

Mr. Valentino can`t go over his budget of $25. The amount of money times the amount of boxes needs to be less than or equal to $25:
Answer: A ) $1.75 x ≤ $25,
where x is the amount of boxes he bought.

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3 years ago

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During the period 1985–2012, the projected enrollment B (in thousands of students) in public schools and the projected enrollmen

Zina [86]

The equation that models the difference in the projected enrollments for public schools and private schools as a function of the number of years since 1985 is B - R = (-18.53t^2 + 975.8t + 48140) - (80.8t + 8049) = -18.53t^2 + 975.8t + 48140 - 80.8t - 8049 = -18.53t^2 + 895t + 40091

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3 years ago

Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,

sweet [91]

Answer:

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

$f'' = x - \frac{3}{2}$

$f' = \int {\left(x-\frac{3}{2}\right) } \, dx$

$f' = \int {x} \, dx -\frac{3}{2}\int \, dx$

$f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C$ , where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative ( $f'(4) = 1$ ):

$1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C$

$C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)$

$C = -1$

The first derivative is $y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1$ , and the particular solution is found by integrating one more time and using the initial condition ( $f(0) = 0$ ):