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svlad2 [7]
3 years ago
5

A total of $30,000 is invested in two corporate bonds that pay 5.5% and 6.25% simple interest. the investor wants an annual inte

rest income of %1800 from the investments. what is the most that can be invested in the 5,5% bond?
Mathematics
1 answer:
lubasha [3.4K]3 years ago
6 0

Answer:

  $10,000

Step-by-step explanation:

Let x represent the amount invested in the 5.5% bond.

For an annual income of $1800, we want ...

  (5.5%)x + (6.25%)(30,000-x) = 1800

  (-0.75%)x + 1875 = 1800

  -0.0075x = -75 . . . . . . . . . subtract 1875

  x = 10,000 . . . . . . . . . . . . . divide by -0.0075

At most $10,000 should be invested at 5.5% to obtain at least $1800 in interest.

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If an arrow is shot upward on Mars with a speed of 68 m/s, its height in meters t seconds later is given by
Talja [164]

Answer:

(a) dy/dt = 68 - 3.72t

(b) 64.28m/s

Step-by-step explanation:

Given y = 68t - 1.86t²

Average Speed is rate of chance of distance with respect to time. Written as dy/dt

To obtain dy/dt, we differentiate y with respect to t, doing that, we have

dy/dt = 68 - 2(1.86)t

= 68 - 3.72t

(a) dy/dt = 68 - 3.72t

(b) To estimate the speed when t = 1, we substitute t = 1 in dy/dt .

dy/dt at t = 1 is

68 - 3.72(1)

= 68 - 3.72

= 64.28

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6 0
3 years ago
Which phrase best describes the translation from the graph y=(x-5)^2+7 to the graph of y=(x+1)^2-2
Helga [31]

Start from the parent function f(x)=x^2


In the first case, you are computing


f(x-5)+7


In the second case, you are computing


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On the other hand, when you transform f(x) \to f(x)+k, you translate the function vertically, k units up if k>0 and k units down if k.


So, the first function is the "original" parabola f(x)=x^2, translated 5 units right and 7 units up. Likewise, the second function is the "original" parabola f(x)=x^2, translated 1 units left and 2 units down.


So, the transformation from (x-5)^2+7 to (x+1)^2-2 is: go 6 units to the left and 2 units down

8 0
3 years ago
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Answer:

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Step-by-step explanation:

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