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Yakvenalex [24]

3 years ago

14

How do you show the work for 1162 divided by 4

1 answer:

Blizzard [7]3 years ago

6 0

You would solve, so 1162/4. Show the steps of you dividing it down on notes of paper

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Pleaseee help me!!! I need the answer!

lora16 [44]

I don’t know how to

3 0

3 years ago

Read 2 more answers

Terrence gets a loan with a $50 processing fee. The loan is for $700 with an interest rate of 8% for one

agasfer [191]

Answer:

15.14%

Step-by-step explanation:

The formula for APR is stated thus:

APR=fees+interest/principal/n*365*100

principal is the loan amount of $700

fees is the processing fees on the loan which is $50

interest amount=principal*interest %=$700*8%=$56

n is the number of days of the loan which is a year i.e 365 days

APR=($50+$56)/$700/365*365*100

APR=$106/$700/365*365*100

APR=0.151428571 /365*365*100

APR=0.151428571 *100=15.14%

The annual percentage rate on the loan is 15.14% which represents the actual cost on the loan not just the interest cost of 8% annually

4 0

3 years ago

Write three different ways for7/10

HACTEHA [7]

The only ways i thought of was 0.7, 0.70, and 0.700 .

7 0

3 years ago

Read 2 more answers

X/(4.2 )= -3 <br><br><br><br><br> Theres no choices

Ksivusya [100]

Answer:

-1.40

Step-by-step explanation:

The answer is -1.40 (negative 1 and 2/5) because....

The number would have to be a mixed number.

The number would also have to be a negative.

5 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