Ask question

Ask question

All categories

2 years ago

15

What is 446,221 rounded to the nearest ten what is the answer

2 answers:

KIM [24]2 years ago

7 0

The answer would be 446,220 because 446,221 the 1 is less than 5 so therefore your answer becomes 446,220

lilavasa [31]2 years ago

4 0

The answer would be 446,220.

You might be interested in

help please!! thank you! will give BRAINLIEST!! *graphs, and just answer question b. and c.! thank youuuu

svlad2 [7]

Answer:

B and C

Step-by-step explanation:

7 0

2 years ago

Please answer the question in the picture below

Elanso [62]

The answer is b because there is a negative in the exponent and the number needs to be a single digit and 54 is double digit. hope this helps!

5 0

3 years ago

What is the value of a? brainliest for right answer

Sholpan [36]

Edit: Nevermind I messed up

4 0

3 years ago

Read 2 more answers

Rewrite the following fractions as division problems. a. 7⁄11 b. 5⁄2 c. 9⁄10 d. 7⁄15

larisa [96]

A. 7/11= 7 ÷ 11
B. 5/2= 5 ÷ 2
C. 9/10= 9 ÷ 10
D. 7/15= 7 ÷ 15

A fraction is just a division problem. 7/11 would be considered 7 ÷ 11 (or just 7/11) just do the same for the rest of the fractions.

3 0

3 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

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

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

So

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

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

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

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

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