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3 years ago

How many times can 21 go into 44

Mathematics

2 answers:

Delvig [45]3 years ago

8 0

Answer:

all work is shown and pictured

adelina 88 [10]3 years ago

3 0

21 can go into 44 2 times

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Helppppp me what is the equation

dem82 [27]

Step-by-step explanation:

can you type the answer 2?

3 0

3 years ago

Read 2 more answers

56. A track coach recorded all runners' times for the 200-meter dash and

ddd [48]

Considering that outliers will be disconsidered, the correct statement is:

The coach can use the Interquartile Range to summarize the runners' times.

<h3>What is the Interquartile Range(IQR) of a data-set?</h3>

It is the values of a data-set that are between the 25th percentile and the 75th percentile, that is, the middle 50% of the values. It's advantage in certain situation is that it disconsiders outliers.

In this problem, the best way to summarize the times is disconsidering outliers, hence the Interquartile Range is used, as it indicates how close all runners' times were to each other.

More can be lerned about the Interquartile Range at brainly.com/question/17083142

6 0

2 years ago

Please I need help with differential equation. Thank you

Inga [223]

1. I suppose the ODE is supposed to be

$\mathrm dt\dfrac{y+y^{1/2}}{1-t}=\mathrm dy(t+1)$

Solving for $\dfrac{\mathrm dy}{\mathrm dt}$ gives

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{y+y^{1/2}}{1-t^2}$

which is undefined when $t=\pm1$ . The interval of validity depends on what your initial value is. In this case, it's $t=-\dfrac12$ , so the largest interval on which a solution can exist is $-1\le t\le1$ .

2. Separating the variables gives

$\dfrac{\mathrm dy}{y+y^{1/2}}=\dfrac{\mathrm dt}{1-t^2}$

Integrate both sides. On the left, we have

$\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\int\frac{\mathrm dz}{z+1}$

where we substituted $z=y^{1/2}$ - or $z^2=y$ - and $2z\,\mathrm dz=\mathrm dy$ - or $\mathrm dz=\dfrac{\mathrm dy}{2y^{1/2}}$ .

$\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\ln|z+1|=2\ln(y^{1/2}+1)$

On the right, we have

$\dfrac1{1-t^2}=\dfrac12\left(\dfrac1{1-t}+\dfrac1{1+t}\right)$

$\displaystyle\int\frac{\mathrm dt}{1-t^2}=\dfrac12(\ln|1-t|+\ln|1+t|)+C=\ln(1-t^2)^{1/2}+C$

$2\ln(y^{1/2}+1)=\ln(1-t^2)^{1/2}+C$

$\ln(y^{1/2}+1)=\dfrac12\ln(1-t^2)^{1/2}+C$

$y^{1/2}+1=e^{\ln(1-t^2)^{1/4}+C}$

$y^{1/2}=C(1-t^2)^{1/4}-1$

I'll leave the solution in this form for now to make solving for $C$ easier. Given that $y\left(-\dfrac12\right)=1$ , we get

$1^{1/2}=C\left(1-\left(-\dfrac12\right)^2\right))^{1/4}-1$

$2=C\left(\dfrac54\right)^{1/4}$

$C=2\left(\dfrac45\right)^{1/4}$

and so our solution is

$\boxed{y(t)=\left(2\left(\dfrac45-\dfrac45t^2\right)^{1/4}-1\right)^2}$

3 0

3 years ago

I need help with Part B.

saveliy_v [14]

Just look at both picture and you are going to understand

4 0

3 years ago

Read 2 more answers

Hey can you please help me posted picture of question

GREYUIT [131]

To solve this problem you must apply the proccedure shown below:

1. You have the following expression:

(3+3i)-(13+15i)

2. If you want to substract both terms, you need to substract the real numbers and the complex numbers. Then, you obtain:

3+3i-13-15i
(3-13)+(3i-15i)

3. Then, you obtain the following result:

-10-12i

4. Therefore, as you can see, the correct answer is the last option (option D), which is:
D. -10-12i

4 0

4 years ago