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3 years ago

9

Kelly is selling magazine subscriptions as part of a school fundraiser during the first 72 days of the fundraiser kelly sold 56

subscriptions you kill me how many subscriptions with kelly sale in 216 day period

1 answer:

Anettt [7]3 years ago

3 0

Kelly sells 52 subscriptions every 72 days, therefore,

216 / 72 = 3.

52 * 3 = 156

Kelly sells 156 subscriptions every 216 days

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4x-3>12x+4<br>Need help solving.

Vika [28.1K]

-3 - 4 > 12 x - 4x

-7 > 8x

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3 years ago

Please help me answer this question

strojnjashka [21]

Answer:

modulus: 3√2
argument: -3π/4 (or 5π/4)

Step-by-step explanation:

The modulus is the magnitude of the complex number; the argument is its angle (usually in radians).

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<h3>rectangular form</h3>

The complex number can be cleared from the denominator by multiplying numerator and denominator by its conjugate:

$\dfrac{-9+3i}{1-2i}=\dfrac{(-9+3i)(1+2i)}{(1-2i)(1+2i)}=\dfrac{-9+3i-18i-6}{1+4}=-3-3i$

<h3>polar form</h3>

The magnitude of this number is the root of the sum of the squares of the real and imaginary parts:

modulus = √((-3)² +(-3)²) = 3√2

The argument is the arctangent of the ratio of the imaginary part to the real part, taking quadrant into consideration.

arg = arctan(-3/-3) = -3π/4 or 5π/4 . . . . radians

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modulus∠argument = (3√2)∠(-3π/4)

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3 years ago

Evaluate the integral of 17/(x^(3)-125)

kvv77 [185]

$x^3-125=(x-5)(x^2+5x+25)$

$\dfrac{17}{(x-5)(x^2+5x+25)}=\dfrac a{x-5}+\dfrac{bx+c}{x^2+5x+25}$
$\dfrac{17}{(x-5)(x^2+5x+25)}=\dfrac{a(x^2+5x+25)+(bx+c)(x-5)}{(x-5)(x^2+5x+25)}$
$\implies17=(a+b)x^2+(5a-5b+c)x+(25a-5c)$
$\implies\begin{cases}a+b=0\\5a-5b+c=0\\25a-5c=17\end{cases}\implies a=\dfrac{17}{75},b=-\dfrac{17}{75},c=-\dfrac{34}{15}$

$\displaystyle\int\dfrac{17}{x^3-125}\,\mathrm dx=\dfrac{17}{75}\int\dfrac{\mathrm dx}{x-5}-\dfrac{17}{75}\int\frac x{x^2+5x+25}\,\mathrm dx-\dfrac{34}{15}\int\dfrac{\mathrm dx}{x^2+5x+25}$
$\displaystyle=\dfrac{17}{75}\int\dfrac{\mathrm dx}{x-5}-\dfrac{17}{150}\int\frac{2x+5}{x^2+5x+25}\,\mathrm dx-\dfrac{17}{10}\int\dfrac{\mathrm dx}{x^2+5x+25}$
$\displaystyle=\dfrac{17}{75}\int\dfrac{\mathrm dx}{x-5}-\dfrac{17}{150}\int\frac{2x+5}{x^2+5x+25}\,\mathrm dx-\dfrac{17}{10}\int\dfrac{\mathrm dx}{\left(x+\frac52\right)^2+\frac{75}4}$

The first integral is trivial. For the second, replace $y=x^2+5x+25$ so that $\mathrm dy=(2x+5)\,\mathrm dx$ . For the third, take $x+\dfrac52=\dfrac{5\sqrt3}2\tan z$ so that $\mathrm dx=\dfrac{5\sqrt3}2\sec^2z\,\mathrm dz$ .

$\displaystyle=\dfrac{17}{75}\ln|x-5|-\dfrac{17}{150}\int\frac{\mathrm dy}y-\dfrac{17}{10}\int\dfrac{\frac{5\sqrt3}2\sec^2z}{\left(\frac{5\sqrt3}2\tan z\right)^2+\frac{475}4}\,\mathrm dz$
$\displaystyle=\dfrac{17}{75}\ln|x-5|-\dfrac{17}{150}\ln|y|-\dfrac{17}{25\sqrt3}\int\dfrac{\sec^2z}{\tan^2z+1}\,\mathrm dz$
$\displaystyle=\dfrac{17}{75}\ln|x-5|-\dfrac{17}{150}\ln(x^2+5x+25)-\dfrac{17}{25\sqrt3}\int\mathrm dz$
$\displaystyle=\dfrac{17}{75}\ln|x-5|-\dfrac{17}{150}\ln(x^2+5x+25)-\dfrac{17}{25\sqrt3}z+C$
$\displaystyle=\dfrac{17}{75}\ln|x-5|-\dfrac{17}{150}\ln(x^2+5x+25)-\dfrac{17}{25\sqrt3}\arctan\dfrac{2x+5}{5\sqrt3}+C$

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3 years ago