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2 years ago

10

A recipe calls for 2 2 over 4 cups of raisins but julie only has a 1 over 4 c up measuring how many 1 over 4 cups does julie nee

ds to measure out2 2 over 4 cups of raisins

1 answer:

Kamila [148]2 years ago

8 0

You will need 6/4 cups of raisins to complete the amount you need to comply with the recipe

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Can someone solve this please?<br><br> 2ab³(2³ a³ b²)

Vilka [71]

I want to say its

6ab12 + 2ab5

4 0

2 years ago

Read 2 more answers

Crawling Crayfish!

boyakko [2]

Answer: Crayfish, 25 s

Step-by-step explanation:

Given

The length of race is 60 cm

Flicker gets a 10 cm head-start

Cray fish covers 6 cm in 5 s i.e. its speed is

$\Rightarrow \dfrac{6}{5}=1.2\ cm/s$

Flicker covers 4 cm in 5 s, its speed is

$\Rightarrow \dfrac{4}{5}=0.8\ cm/s$

Time taken by Cray fish to cover the race is

$\Rightarrow t_1=\dfrac{60}{1.2}\\\\\Rightarrow t_1=50\ s$

Time taken by flicker to cover race with 10 cm head start

$\Rightarrow t_2=\dfrac{60-10}{0.8}\\\\\Rightarrow t_2=62.5\ s$

Time taken by crayfish is less. Hence, crayfish wins the race

When they both covers the same distance, they tied momentarily i.e.

$\Rightarrow 1.2t=10+0.8t\\\Rightarrow 0.4t=10\\\\\Rightarrow t=\dfrac{10}{0.4}\\\\\Rightarrow t=25\ s$

After 25 s, they tied the race.

5 0

2 years ago

An educator believes that new reading activities for elementary school children will improve reading

Damm [24]

The correct answer to this open question is the following.

Unfortunately, you forgot to attach the scores shown in the back-to-back stem-and-leaf display. So we do not know what the numbers are and we do not have any reference at all

What we can do to help you is to comment on the following general terms.

There have been previous and similar experiments or projects like this in other schools in America. These results suggest that the new activities are better because extra or special reading comprehension programs better prepare students to understand what they are reading and comprehend more than basic ideas of the text.

Students that participate in these programs develop a better sense to understand and like what they are reading, considerably increasing their focus and attention.

These programs have resulted positively when trying to improve the marks of the students, compared to other traditional approaches.

6 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

2 years ago

Llana [10]

Answer:

Option D is not true

Step-by-step explanation:

Whole Numbers (W) = They include Natural numbers (N) and 0.

7 0

2 years ago