How much would $300 invested at 7% interest compounded continuously be

Mathematics

2 answers:

Paha777 [63]2 years ago

4 0

The correct awnser would be B because it is the only one rounded to the nearest cent

katen-ka-za [31]2 years ago

4 0

The answer would be B

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At noon, ship a is 30 nautical miles due west of ship

Dmitry_Shevchenko [17]

Let the origin of the coordinate system be the position of ship B at noon. Then the distance between the ships as a function of t (in hours) is
.. d = √((30 +23t)² +(17t)²)
.. = √(818t² +1380t +900)

The rate of change of distance with respect to time is
.. d' = (1/2)(2*818*t +1380)/√(818t² +1380t +900)
.. = (818t +690)/√(818t² +1380t +900)

At t=4, this is
.. d' = (818*4 +690)/√(818*16 +1380*4 +900)
.. = 3962/√19508
.. ≈ 28.37 . . . . . knots

The distance between the ships is increasing at about 28.37 knots at 4 pm.

8 0

2 years ago

Simplify the expression.<br><br> (−5g5h6)2(g4h2)4

Colt1911 [192]

We have to simplify the expression here. The expression given,

$(-5g^5h^6)^2(g^4h^2)^4$

Now to find $(-5g^5h^6)^2$ we will use power of a product property. We have to distribute the power here. We will get,

$(-5g^5h^6)^2 = (-5)^2(g^5)^2(h^6)^2$

= $25(g^5)^2(h^6)^2$

Now we will use power of a power property. If given $(a^m)^n$ we will have to multtiply the powers and we will get $a^{mn}$ . So we will get here,

$25(g^5)^2(h^6)^2 = 25 g^{10} h^{12}$

Now the next part is $(g^4h^2)^4$

By using power of a power property we will get,

$(g^4)^4(h^2)^4 = g^{16}h^8$

Now we have to multiply them.

$(25g^{10}h^{12} )(g^{16}h^8)$

Now we have to use product of powers property. If we have same base them we will have to add the exponents there. The formula is $(a^m)(a^n) = a^{(m+n)}$ . So we will get here,