If 10 cups of flour is needed to make 3 loaves of bread, how many cups of flour are needed to make 17 loaves?

Find the slope of the line that passes through points (3,5) and (8,5)

Katyanochek1 [597]

Answer: m = 0

Step-by-step explanation: To solve this problem, we don't even have to use our slope formula. It's important to understand that when we have the same y coordinate in our first ordered pair and our second ordered pair, this means that the line will be flat or horizontal.

When a line is horizontal, it means the line has a slope of zero.

We use the variable m to represent slope.

So here, we can say that m = 0.

4 0

3 years ago

a researcher wishes to design a fully blocked experiment with groups of subjects representing every possible combination of 1 ex

lesantik [10]

You have 6 variables with 2 levels each. Thus 2*2*2*2*2*2 or 2^6 or 64.

5 0

3 years ago

Read 2 more answers

What is a factor of 21x² + 13x - 20

Tems11 [23]

Answer:

(7x-5)(3x+4)

Step-by-step explanation:

4 0

3 years ago

Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions

monitta

I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take $z=y'$ , so that $z'=y''$ and we're left with the ODE linear in $z$ :

$y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2$

Now suppose $y$ has a power series expansion

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Then the ODE can be written as

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0$

$\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0$

All the coefficients of the series vanish, and setting $x=0$ in the power series forms for $y$ and $y'$ tell us that $y(0)=a_0$ and $y'(0)=a_1$ , so we get the recurrence

$\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}$

We can solve explicitly for $a_n$ quite easily:

$a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}$

and so on. Continuing in this way we end up with

$a_n=\dfrac{a_1}{n!}$

so that the solution to the ODE is

$y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x$

We also require the solution to satisfy $y(0)=a_0$ , which we can do easily by adding and subtracting a constant as needed:

$y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}$

4 0

3 years ago

1.In a class of 45 pupils ,21play chess and 23 play cards and 5play both games

Mumz [18]

Step-by-step explanation:

(a)

ii.

21 - 5 = 16 play only chess.

iii.

23 - 5 = 18 play only cards

5 play both.

that is 16+18+5 = 39 that play at least one of these games.

i.

so, 45 - 39 = 6 don't play any of these games.

(b)

we have 16+18 = 34 pupils out of 45 that play exactly one game.

so the probabilty to pick one of them is as usual

desired cases / total cases = 34/45 =

= 0.755555555... ≈ 0.76

8 0

2 years ago