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harina [27]
3 years ago
9

I NEED HELP ASAP !!!!

Mathematics
1 answer:
mestny [16]3 years ago
8 0
D would me the answer I just took the assignment
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A sum amount <br>to rupees 34476 in two and a half years at 4% CI. the sum is ​
Kruka [31]

Given:

Amount = Rs. 34476

Rate of compound interest = 4%

Time = 2\dfrac{1}{2}=2.5 years

To find:

The principal value.

Solution:

Formula for amount is

A=P\left(1+\dfrac{r}{100}\right)^t

Where, P is principal value, r is rate of interest and t is time in years.

Putting the given values, we  get

34476=P\left(1+\dfrac{4}{100}\right)^{2.5}

34476=P\left(1+0.04\right)^{2.5}

34476=P\left(1.04\right)^{2.5}

\dfrac{34476}{\left(1.04\right)^{2.5}}=P

Now,

P=31256.0090

P\approx 31256

Therefore, the value of sum or principal value is Rs.31256.

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2 years ago
These Are My Answers, Are They Correct ?
prisoha [69]
Haha number one is 3 and two is 2
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2 years ago
Rewrite the quadratic function below in standard form y=3(x+2)(x-3)
babunello [35]
X=3(x^2-x_6)
3x^2-x-6
8 0
2 years ago
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Find parametric equations for the line through the point (0, 2, 2) that is parallel to the plane x + y + z = 4 and perpendicular
Oxana [17]

Answer: X = 3t, Y =2 - t, Z =2

Step-by-step explanation: the plane

x + y + z =4has normal vector

M =<1,1,1> and the line

x = 1 + t, y = 2 − t, z = 2t has direction

v =<1, −1, 2>. So the vector

A= n × v

=<1, 1, 1> × <1, −1, 2>

=<2−(−1),1−2,−1−1>

=<3,−1,−2>

5 0
3 years ago
Find the following integral
ololo11 [35]

There's nothing preventing us from computing one integral at a time:

\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2

\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2

\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx

Expand the integrand completely:

x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x

Then

\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}

4 0
2 years ago
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