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2 years ago

Ella and Amy are planning trips to

Mathematics

1 answer:

Vsevolod [243]2 years ago

5 0

Applying the combination formula, the number of ways there are is: 66.

<h3>What is Combination?</h3>

Combination is a technique in maths used in selecting objects from a group of objects in a way that the order in which they are selected does not matter.

Combination formula is given as:

$nC_r = \frac{n!}{r(n - r)!}$

Given the following:

n = 12
r = 2

Plug in the values:

$12C_2 = \frac{12!}{2(12 - 2)!}\\\\12C_2 = \frac{12!}{2(10)!}\\\\12C_2 = \frac{12 \times 11}{2(1)}$

$\mathbf{12C_2 = 66}$

Therefore, applying the combination formula, the number of ways there are is: 66.

Learn more about combination on:

brainly.com/question/25821700

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How do u find percent of change?

Ksju [112]

Answer:

You find percent of change

by finding the differences between the numbers given

and by setting up a fraction to find the change because percentages deals with fractions

Then you simplify ur fraction by multiplying by 100

Step-by-step explanation:

e.g

find the percent of change from 20 to 80

find their differences

80-20= 60

set up a fraction

60/20

simplify your fraction by multiplying over 100

60/20*100

3*100=300%. answer

7 0

3 years ago

Answer the following questions CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!

QveST [7]

Answer by JKismyhusbandbae: This is not a function because the two in the domain area is paired with 21 and 25. A function is basically whenever there is no domain that repeats.

8 0

4 years ago

Read 2 more answers

Find the volume of the hemisphere

seraphim [82]

Answer:

0.341 $\pi$ or 1.0723

Step-by-step explanation:

the volume of hemisphere = (2/3) $\pi$ $r^{3}$

r = (1/2)d = 0.8

(2/3) $\pi$ $(0.8)^{3}$ = 0.341 $\pi$ = 1.0723

3 0

3 years ago

Solve the initial value problem 2ty" + 10ty' + 8y = 0, for t > 0, y(1) = 1, y'(1) = 0.

Eva8 [605]

I think you meant to write

$2t^2y''+10ty'+8y=0$

which is an ODE of Cauchy-Euler type. Let $y=t^m$ . Then

$y'=mt^{m-1}$

$y''=m(m-1)t^{m-2}$

Substituting $y$ and its derivatives into the ODE gives

$2m(m-1)t^m+10mt^m+8t^m=0$

Divide through by $t^m$ , which we can do because $t\neq0$ :

$2m(m-1)+10m+8=2m^2+8m+8=2(m+2)^2=0\implies m=-2$

Since this root has multiplicity 2, we get the characteristic solution

$y_c=C_1t^{-2}+C_2t^{-2}\ln t$

If you're not sure where the logarithm comes from, scroll to the bottom for a bit more in-depth explanation.

With the given initial values, we find

$y(1)=1\implies1=C_1$

$y'(1)=0\implies0=-2C_1+C_2\implies C_2=2$

so that the particular solution is

$\boxed{y(t)=t^{-2}+2t^{-2}\ln t}$

# # #

Under the hood, we're actually substituting $t=e^u$ , so that $u=\ln t$ . When we do this, we need to account for the derivative of $y$ wrt the new variable $u$ . By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm dy}{\mathrm du}$

Since $\frac{\mathrm dy}{\mathrm dt}$ is a function of $t$ , we can treat $\frac{\mathrm dy}{\mathrm du}$ in the same way, so denote this by $f(t)$ . By the quotient rule,

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac ft\right]=\dfrac{t\frac{\mathrm df}{\mathrm dt}-f}{t^2}$

and by the chain rule,

$\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\mathrm df}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm df}{\mathrm du}$

where

$\dfrac{\mathrm df}{\mathrm du}=\dfrac{\mathrm d}{\mathrm du}\left[\dfrac{\mathrm dy}{\mathrm du}\right]=\dfrac{\mathrm d^2y}{\mathrm du^2}$

so that

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm dy}{\mathrm du}}{t^2}=\dfrac1{t^2}\left(\dfrac{\mathrm d^2y}{\mathrm du^2}-\dfrac{\mathrm dy}{\mathrm du}\right)$

Plug all this into the original ODE to get a new one that is linear in $u$ with constant coefficients:

$2t^2\left(\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm d y}{\mathrm du}}{t^2}\right)+10t\left(\dfrac{\frac{\mathrm dy}{\mathrm du}}t\right)+8y=0$

$2y''+8y'+8y=0$

which has characteristic equation

$2r^2+8r+8=2(r+2)^2=0$

and admits the characteristic solution

$y_c(u)=C_1e^{-2u}+C_2ue^{-2u}$

Finally replace $u=\ln t$ to get the solution we found earlier,

$y_c(t)=C_1t^{-2}+C_2t^{-2}\ln t$

4 0

4 years ago

How many real solutions if any, does 2x^2-3x+8=0

fenix001 [56]

Answer:

The quadratic equation $2\, x^{2} - 3\, x + 8 = 0$ has no real solution.

Step-by-step explanation:

Rewrite the quadratic equation $2\, x^{2} - 3\, x + 8 = 0$ in the standard form $a\, x^{2} + b\, x + c = 0$ :

$2\, x^{2} + (-3)\, x + 8 = 0$ , for which:

$a = 2$ .
$b = (-3)$ .
$c = 8$ .

The quadratic discriminant of $a\, x^{2} + b\, x + c = 0$ is $(b^{2} - 4\, a\, c)$ . The quadratic discriminant of $2\, x^{2} + (-3)\, x + 8 = 0$ would be:

$\begin{aligned}& b^{2} - 4\, a\, c \\ =\; & (-3)^{2} - 4 \times 2 \times 8 \\ =\; & (-55)\end{aligned}$ .

Since the quadratic discriminant of this equation is negative, this quadratic equation has no real solution.

3 0

2 years ago