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3 years ago

What will be the compound interest p=25000 r= 12% p.a for 5 years

Mathematics

2 answers:

DanielleElmas [232]3 years ago

6 0

Answer:

Step-by-step explanation:

I= PRT ÷ 100

P is principal

R is rate (%)

T is time (years)

I = (25000 × 12 × 5) ÷ 100

= 15000

rusak2 [61]3 years ago

5 0

Answer:19057.5

Step-by-step explanation:

Compound interest=p(1+r/100)^n-p

Compound interest=25000(1+12/100)^5-25000

Compound interest=25000(1.12)^5-25000

Compound interest=25000×1.7623-25000

Compound interest=44057.5-25000

Compound interest=19057.5

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The length of the missing side is: ?m

victus00 [196]

Answer:

85 ft

Step-by-step explanation:

for all right triangles, triangles with a 90 degree angle, the sides are equal to $a^{2} + b^{2} = c^{2}$ where c is the hypotenuse, or the longest side.

so... $40^{2} + 75^{2} = c^{2}$

$7225 = c^{2}$

c= 85

6 0

2 years ago

HELP!!!<br> HALP MEEEEEEEE

goldfiish [28.3K]

Answer:

Bryce's unit rate in miles around the track is 3 rate per mile . Now, as we know the formula for speed i.e. So, Bryce's unit rate in miles around the track is 3 rate per mile .

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3 years ago

Read 2 more answers

Y= (x-9) (x+3)

juin [17]

Answer:

y = x² − 6x − 27

Step-by-step explanation:

To distribute, you can use something called FOIL. It stands for First, Outer, Inner, Last.

First, multiply the First term in each factor.

x · x = x²

Now multiply the Outer terms in each factor.

x · 3 = 3x

Next multiply the Inner terms in each factor.

-9 · x = -9x

Finally, multiply the Last terms in each factor.

-9 · 3 = -27

Add them all up:

x² + 3x − 9x − 27

x² − 6x − 27

4 0

3 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago

Help me please this is a quiz for today

Alexandra [31]

Slope is 5
Y intersect is 0

Y= 5x

3 0

3 years ago