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3 years ago

7

Convert the following numbers in to quinary numeration system 7521

2 answers:

mel-nik [20]3 years ago

7 0

220041 is the answer

Julli [10]3 years ago

4 0

Answer:

$220041_{5}$

Step-by-step explanation:

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Find the other endpoint of the line segment with the given endpoint and midpoint.

Shtirlitz [24]

Answer:

endpoint is (29, -13)

Step-by-step explanation:

Formula for midpoint is $(\frac{x_{1} + x_{2} }{2} ,\frac{y_{1} + y_{2} }{2} )$

So, let $(x_{2} , y_{2} )$ be the coordinates of the endpoint you are looking for.

$x_{1} = -9$ and $y_{1} = 7$

$(\frac{-9 + x_{2} }{2} , \frac{7 + y_{2} }{2} )$ = (10, -3)

$\frac{-9 + x_{2} }{2} = 10$ $\frac{7 + y_{2} }{2} = -3$

-9 + $x_{2}$ = 20 7 + $y_{2} = -6$

$x_{2} = 29$ $y_{2} = -13$

endpoint is (29, -13)

6 0

3 years ago

The perimeter of a rectangle is 24 centimeters.Its length is 9 centimeters. Find the width

34kurt

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Share £90 in the ratio 1:5<br> HELP ME PLEASE

ZanzabumX [31]

Answer:

the answer is 15:75

Step-by-step explanation:

(1+5)=6

£90/6=15

So, (1×15):(5×15)

=15:75

7 0

3 years ago

Three friends go bowling the cost per person per game is $ 5.30 the cost to rent shoes is $ 2.50 per person.Their total cost is

siniylev [52]

Find cost of shoes first (7.50)
subtract from the total value (55.20-7.50= 47.7)
then divide by 5.30 (47.7/ 5.3= 9)

4 0

3 years ago

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

Stells [14]

$y'=(t+y)^2-1$

Substitute $u=t+y$ , so that $u'=y'$ , and

$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

3 0

3 years ago