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3 years ago

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Which situation can be represented by the equation y = 12x? A) Victoria went to school for x years. This is 12 times y, the numb

er of years her brother went to school. B) Victoria spent x dollars to buy a gift for her brother. She gave the cashier y dollars and received $12 in change. C) Victoria has y dollars. This amount is 12 times x, the amount of money in dollars Victoria’s brother has. D) Victoria is y years old. Her age is 12 years greater than x, her brother’s age in years.

1 answer:

dolphi86 [110]3 years ago

6 0

C) Victoria has y dollars. This amount is 12 times x, the amount of money in dollars Victoria’s brother has.

<u>Step-by-step explanation</u>:

The equation is represented as y = 12x

The option A) does not match the equation. since it is given that x is 12 times y (x = 12y).
The option B) does not match the equation since it is given that she recieves $12 as change (x = y-12)
The option C) matches the equation since Victoria has 12 times the amount of his brother (y = 12x)
The option D) does not match the equation since Victoria is 12 years greater (y = x+12)

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Answer:

She have to pay the bank at the end of the 6 years = 222000$

Step-by-step explanation:

Formula applied in this case where price, interest rate and time duration is mentioned.

Interest = prt/100

Substituting all the given values in the formula.

interest = 15000 * 6 * 8/100

150 * 6 * 8 = 7200

Interest + 15000 = 15000 + 7200 = 222000$ has to pay the total.

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Answer:

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Step-by-step explanation:

Total number of people, x=300

Number of people donated blood with type A, y

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Number of people donated blood with type B, z

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Solution for dy/dx+xsin 2 y=x^3 cos^2y

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Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
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Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
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So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
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Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
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$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
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You should end up with

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