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3 years ago

What is the annual percentage yield(APY) for money invested at an annual rate of

Mathematics

1 answer:

Lady_Fox [76]3 years ago

8 0

A.) 5.15% compounded continuously?

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it is exponent less than 1

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The probability that a seventh grader chosen at random will play an instrument other than bass is 72%.

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the answer will be 10 aka number 1

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If the left side is 18 and the number above it is half that aka 9 then the other side is 10

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2 years ago

What is the radius of the circle defined by x2 + 2x + y2 + 6y − 39 = 0? 49 9 7 81

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radius = 7

the equation of a circle in standard form is

(x-a)² + (y-b)² = r²

where (a, b) are the coordinates of the centre and r is the radius

To obtain this form ' complete the square ' on both the x and y terms

x² + 2(1)x + 1 + y² + 2(3)y + 9 = 39 + 1 + 9

(x+1)² + (y+3)² = 49 ← in standard form

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3 years ago

Read 2 more answers

A homeowner plans to enclose a 200 square foot rectangular playground in his garden, with one side along the boundary of his pro

Anestetic [448]

Answer:

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

Step-by-step explanation:

Given that, a homeowner plans to enclose a 200 square foot rectangle playground.

Let the width of the playground be y and the length of the playground be x which is the side along the boundary.

The perimeter of the playground is = 2(length +width)

=2(x+y) foot

The material costs $1 per foot.

Therefore total cost to give boundary of the play ground

=$[ 2(x+y)×1]

=$[2(x+y)]

But the neighbor will play one third of the side x foot.

So the neighbor will play $=\$(\frac13x)$

Now homeowner's total cost for the material is

$=\$[ 2(x+y)-\frac13x]$

$=\$[2x+2y-\frac13x]$

$=\$[2y+x+x-\frac13x]$

$=\$[2y+x+\frac{3x-x}{3}]$

$=\$[2y+x+\frac{2}{3}x]$

$=\$[2y+\frac53x]$

$\therefore C(x)=2y+\frac53x$

where C(x) is total cost of material in $.

Given that the area of the playground is 200.

We know that,

The area of a rectangle is =length×width

=xy square foot

∴xy=200

$\Rightarrow y=\frac{200}{x}$

Putting the value of y in C(x)

$\therefore C(x)=2(\frac{200}{x})+\frac53x$

The domain of C is $(0,\infty )$ .

$\therefore C(x)=2(\frac{200}{x})+\frac53x$

Differentiating with respect to x

$C'(x)= - \frac{400}{x^2}+\frac53$

Again differentiating with respect to x

$C''(x) = \frac{800}{x^3}$

To find the critical point set C'(x)=0

$\therefore 0= - \frac{400}{x^2}+\frac53$

$\Rightarrow \frac{400}{x^2}=\frac{5}{3}$

$\Rightarrow x^2 =\frac{400\times 3}{5}$

$\Rightarrow x=\sqrt{240}$

$\Rightarrow x=15.49 \approx15.5$

Therefore

$\left C''(x) \right|_{x=15.5}=\frac{800}{15.5^3}>0$

Therefore at x= 15.5 , C(x) is minimum.

Putting the value of x in $y=\frac{200}{x}$ we get

$\therefore y=\frac{200}{15.5}$

=12.9

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

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3 years ago