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Tresset [83]
3 years ago
15

how much would $500 invested at 6% interest compounded continuously be worth after 4 years? round your answer to the nearest cen

t
Mathematics
2 answers:
kirill115 [55]3 years ago
6 0

Answer:

He would worth $ 635.62 ( approx )

Step-by-step explanation:

Since, the amount compounded continuously,

A=Pe^{rt}

Where, P is the principal amount,

R is the rate per period,

t is the number of periods,

Here, P = $ 500,

r = 6 % = 0.06,

t = 4 years,

By substituting the values,

The amount after 4 years would be,

A=500e^{0.06\times 4}

=500e^{0.24}

=\$635.624575161\approx \$635.62

Mashcka [7]3 years ago
3 0
A=pe^rt
A=500×e^(0.06×4)
A=635.62
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A textbook store sold a combined total of 426 math and sociology textbooks in a week. The number of math textbooks sold was 88 m
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Let m and s be the number of math and sociology books sold respectively.

m=s+88 we are also told that:

m+s=426, using m found above in this equation gives you:

s+88+s=426 combine like terms on left side

2s+88=426  subtract 88 from both sides

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2 years ago
Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)

We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>

The vector product pq x pr gives a vector perpendicular to both pq and pr.  This vector is the normal vector of a plane passing through all three points
pq x pr
=
  i   j   k
-4 -2 -4
-3  5  1
=<-2+20,12+4,-20-6>
=<18,16,-26>

Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>

The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).

The equation of the required plane is therefore
Π :  9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π  :  9x+8y-13z=24

Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.

3 0
3 years ago
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