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2 years ago

How many solutions are there to an inequality?

Mathematics

1 answer:

zhenek [66]2 years ago

6 0

Answer: 3

Step-by-step explanation: There can be none, one, or infinite solutions. Therefore there can be up to three possible solutions to an inequality.

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The temperature changed from 7°f at 6pm to -5°f at midnight. What was the difference between the high an low temperatures? Whats

Ronch [10]

Answer:

Difference between high and low temperatures: 12°F

Average change in temperature per hour: 2°F

Step-by-step explanation:

The difference between high and low temperatures was: 7 - (-5) = 12 (°F)

Midnight is 12am, so 6pm to midnight is 6 hours.

The average change in temperature per hour is: 12 / 6 = 2 (°F)

7 0

3 years ago

What is the equation of the asymptote of the graph of y=2(6^x) ?

Arlecino [84]

The value of can never be 0

y = 0

A

6 0

3 years ago

Slove : 28/36 = 14/y

Dafna1 [17]

Answer:

y = 18

Step-by-step explanation:

<h3>i) reduce the fraction with 4</h3>

$\frac{28}{36} = \frac{14}{y}$

$\frac{28 \div 4}{36 \div 4} = \frac{14}{y}$

$\frac{7}{9} = \frac{14}{y}$

<h3>ii) simplify the equation with cross multiplication</h3>

$\frac{7}{9} = \frac{14}{y}$

$7y = 14 \times 9$

$7y = 126$

<h3>iii) divide both sides of the equation by 7</h3>

$\frac{7y}{7} = \frac{126}{7}$

$y = 18$

7 0

2 years ago

Read 2 more answers

PLEASE HELP ME I ONLY HAVE 20 MINUTES!!!!

Over [174]

$h(x) = 40x - 16x^2\\h(x) = 15\\40x - 16x^2 = 15\\-16x^2 + 40x - 15 = 0\\$

Using the quadratic formula we get

$x_1 = 0.5\\x_2 = 2.04$

Therefore the tennis ball is at least 15 feet above the ground from 0.5 seconds to 2.04 seconds

7 0

2 years ago

Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7

Zolol [24]

$f(x)=\begin{cases}1&\text{for }-7$

The Fourier series expansion of $f(x)$ is given by

$\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7$

where we have

$a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx$
$a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)$
$a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2$

The coefficients of the cosine series are

$a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx$
$a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)$
$a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}$
$a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}$

When $n$ is even, the numerator vanishes, so we consider odd $n$ , i.e. $n=2k-1$ for $k\in\mathbb N$ , leaving us with

$a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}$

Meanwhile, the coefficients of the sine series are given by

$b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx$
$b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)$
$b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{7(-1)^{n+1}}{n\pi}$

So the Fourier series expansion for $f(x)$ is

$f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7$

3 0

2 years ago