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3 years ago

6

last year a school raised $19,950 by selling $15 tickets to a carnival fundraiser. this year the school will raise $20 per ticke

t. if the school sells the same number of tickets as it did last year, how much money will be raised?

1 answer:

AURORKA [14]3 years ago

7 0

Answer:

$26,600

Step-by-step explanation:

$19,950 / $15 = 1330 tickets

1330 x $20 = 26,600

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A right prism with rhombus bases is shown. the side length of each rhombus is 5 units. the height of the prism is 16 units. the

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3 years ago

Read 2 more answers

2x2 – 5x + 67 = 0<br> What would be your first step in completing the square for the equation above?

7nadin3 [17]

Answer:

The first step is to divide all the terms by the coefficient of $x^{2}$ which is 2.

The solutions to the quadratic equation $2x^2\:-\:5x\:+\:67\:=\:0$ are:

$x=\frac{5}{4}+i\frac{\sqrt{511}}{4},\:x=\frac{5}{4}-i\frac{\sqrt{511}}{4}$

Step-by-step explanation:

Considering the equation

$2x^2\:-\:5x\:+\:67\:=\:0$

The first step is to divide all the terms by the coefficient of $x^{2}$ which is 2.

so

$\frac{2x^2-5x}{2}=\frac{-67}{2}$

$x^2-\frac{5x}{2}=-\frac{67}{2}$

Lets now solve the equation by completeing the remaining steps

Write equation in the form: $x^2+2ax+a^2=\left(x+a\right)^2$

Solving for $a$ ,

$2ax=-\frac{5}{2}x$

$a=-\frac{5}{4}$

$\mathrm{Add\:}a^2=\left(-\frac{5}{4}\right)^2\mathrm{\:to\:both\:sides}$

$x^2-\frac{5x}{2}+\left(-\frac{5}{4}\right)^2=-\frac{67}{2}+\left(-\frac{5}{4}\right)^2$

$x^2-\frac{5x}{2}+\left(-\frac{5}{4}\right)^2=-\frac{511}{16}$

Completing the square

$\left(x-\frac{5}{4}\right)^2=-\frac{511}{16}$

Since, you had required to know the first step in completing the square for the equation above, I hope you have got the point, but let me quickly solve the remaining solution.

For $f^2\left(x\right)=a$ the solution are $f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

Solving

$x-\frac{5}{4}=\sqrt{-\frac{511}{16}}$

$x-\frac{5}{4}=\sqrt{-1}\sqrt{\frac{511}{16}}$

$x-\frac{5}{4}=i\sqrt{\frac{511}{16}}$ ∵ Applying imaginary number rule $\sqrt{-1}=i$

$x-\frac{5}{4}=i\frac{\sqrt{511}}{\sqrt{16}}$

$-\frac{5}{4}=i\frac{\sqrt{511}}{4}$

$x=\frac{5}{4}+i\frac{\sqrt{511}}{4}$

Similarly, solving

$x-\frac{5}{4}=-\sqrt{-\frac{511}{16}}$

$x-\frac{5}{4}=-i\frac{\sqrt{511}}{4}$ ∵ Applying imaginary number rule $\sqrt{-1}=i$

$x=\frac{5}{4}-i\frac{\sqrt{511}}{4}$

Therefore, the solutions to the quadratic equation are:

$x=\frac{5}{4}+i\frac{\sqrt{511}}{4},\:x=\frac{5}{4}-i\frac{\sqrt{511}}{4}$

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2 years ago

A car has a mass of 2400 kg, correct to the nearest hundred kilograms.

erma4kov [3.2K]

Answer:

3805kg

Step-by-step explanation:

2400 ≥ 2350

1460 ≥ 1455

2400 + 1460≥ 2350 + 1455 = 3805

The lower bound is 3805kg

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2 years ago