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3 years ago

What is the greatest common factor for 24 and 16

Mathematics

2 answers:

ludmilkaskok [199]3 years ago

6 0

The greatest common factor is 8

alina1380 [7]3 years ago

4 0

8 is the answer because it is the biggest answer that can fit into both 16 and 24.

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P is inversely proportional to Q.

Volgvan

$If\ P\ is\ inversely\ proportional\ to\ Q\ then:\\\\P\times Q=constans\\\\P=100;\ Q=32\\\\100\times32=3200\\\\P=2Q\\\\therefore\\\\P\times Q=2Q\times Q=2Q^2\\\\2Q^2=3200\ \ \ \ |divide\ both\ sides\ by\ 2\\Q^2=1600\ \ \ \ \ |sqare\ rot\ both\ sides\\Q=\sqrt{1600}\\\boxed{Q=40}$

7 0

3 years ago

Helpppppp me please

motikmotik

I believe it’s 100 hopefully it helps

6 0

4 years ago

Can anyone help me solve this Algebra 2 Problem, i literally cant figure it out

Andrews [41]

Answer:

Zeros : 1 , -1, 3

Degree : 4

End Behaviour : At x-> ∞ f(x) -> ∞ and x->-∞ f(x) -> ∞

Y - intercept : -3

Extra Points: (0,-3), (2,-3)

Step-by-step explanation:

f(x) = 0 to find the zeros

$Therefore (x+1)(x-1)^{2} (x-3) = 0$

Clearly x = -1,1,3

Here 1 is a repeating root as it is (x-1)²

Degree is highest power of x in f(x)

Clearly it is x*x²*x = x⁴ is the maximum power of x

Thus degree is 4

Looking at end behavior we substitute x->∞ and x-> -∞

Clearly f(x)>0 as all terms are positive and f(x)->∞

Similarly when x->-∞

f(x)>0 as 2 terms are -ve and their product is positive thus f(x)-> ∞

Y-Intercept is f(0)

f(0) = (0+1)(0-1)²(0-3) = 1*1*-3 = -3

Thus Y-Intercept is -3

Substitute x = 0 , 2 for extra points

Thus f(0) = -3

and f(2) = -3

Thus points on the graph (0,-3), (0,2)

We can use all this information to draw a graph remember that 1 is a repeating root so that will be a point of minima. The graph is a parabola that passes through x-axis at x = -1, 3.

5 0

3 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

3 years ago

Write the ratio as a fraction in simplest form. 32:42

maks197457 [2]

16/21 is your answer

You can take a 2 out of both numbers to get your answer

I hope this helped! :)

3 0

3 years ago

Read 2 more answers