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Ad libitum [116K]

3 years ago

6

Ted is making a bread recipe that uses 3 1/4 cups of flour and a muffin recipe that uses 2 3/4 cups of flour. how much more flou

r is in the bread than in the muffins? how much flour does ted need for both recipes?

1 answer:

bonufazy [111]3 years ago

6 0

The bread has 1/4 cup more flour

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Solve the equation 2x^2-9x=0 by factoring

qwelly [4]

Answer: x = 0, $\frac{9}{2}$

Step-by-step explanation:

Given:

2x^2-9x=0

Rewrite:

0 = 2x² - 9x

Factor:

0 = (2x - 9)x

Solve:

x = 0, $\frac{9}{2}$

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2 years ago

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Solve the following system of equations. What is the y value of the solution?

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Answer:

y = 2

Step-by-step explanation:

x + 5y - 10 = 0

X = 4y - 8

since x is given as = 4y - 8 we can use this in first equation

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9y - 18 = 0 add 18 to both sides of the equation

9y = 18 divide both sides by 9

y = 2

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4 years ago

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The weight of 10,000 identical samples of a substance is 100 pounds. What is the weight of 10 samples?

Misha Larkins [42]

The answer to the problem is .1

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4 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

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

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Solve the equation 1/3 + 1/4

Ronch [10]

1/3+1/4 = 4/12 + 3/12 = 7/12

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