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mr Goodwill [35]
3 years ago
5

Mr. Guny deposits $4,900 in a savings account that pays 3 1/2% interest compounded quarterly.

Mathematics
1 answer:
Fittoniya [83]3 years ago
7 0

9514 1404 393

Answer:

  see attached

Step-by-step explanation:

A spreadsheet performs these computations nicely. Each quarter's interest is 0.035/4 times the balance at the end of the previous quarter, rounded to the nearest penny. Each quarter's balance is the sum of the interest and the previous balance.

See the attached for the numbers requested.

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The variable z is directly proportional to x, and inversely proportional to y. When x is 20 and y is 6, z has the value 66.66666
sergejj [24]

Answer:

41.667 to the nearest thousandth.

Step-by-step explanation:

z = kx/y where k is the constant of proportionality.

Inserting the given values:

66.666666666667 = k*20/6

20k = 6*66.666666666667

k = 6*66.666666666667 / 20

=  20

So the equation of variation is z = 20x /y

When x = 25 and y = 12:

z = 20*25/12

= 41.666666666667.

6 0
3 years ago
Pretty Pavers company is installing a driveway. Below is a diagram of the driveway they are
prohojiy [21]

Answer:

The most correct option is;

(B) 958.2 ft.²

Step-by-step explanation:

From the question, the dimension of each square = 3 ft.²

Therefore, the length of the sides of the square = √3 ft.

Based on the above dimensions, the dimension of the small semicircle is found by counting the number of square sides ti subtends as follows;

The dimension of the diameter of the small semicircle = 10·√3

Radius of the small semicircle = Diameter/2 = 10·√3/2 = 5·√3

Area of the small semicircle = (π·r²)/2 = (π×(5·√3)²)/2 = 117.81 ft.²

Similarly;

The dimension of the diameter of the large semicircle = 10·√3 + 2 × 6 × √3

∴ The dimension of the diameter of the large semicircle = 22·√3

Radius of the large semicircle = Diameter/2 = 22·√3/2 = 11·√3

Area of the large semicircle = (π·r²)/2 = (π×(11·√3)²)/2 = 570.2 ft.²

Area of rectangle = 11·√3 × 17·√3 = 561

Area, A of large semicircle cutting into the rectangle is found as follows;

A_{(segment \, of \, semicircle)} = \frac{1}{4} \times (\theta - sin\theta) \times r^2

Where:

\theta = 2\times tan^{-1}( \frac{The \, number \, of  \, vertical  \, squrare  \, sides  \ cut  \,  by  \  the  \  large  \,  semicircle}{The \, number \, of  \, horizontal \, squrare  \, sides  \ cut  \,  by  \  the  \  large  \,  semicircle} )

\therefore \theta = 2\times tan^{-1}( \frac{10\cdot \sqrt{3} }{5\cdot \sqrt{3}} ) = 2.214

Hence;

A_{(segment \, of \, semicircle)} = \frac{1}{4} \times (2.214 - sin2.214) \times (11\cdot\sqrt{3} )^2 = 128.3 \, ft^2

Therefore; t

The area covered by the pavers = 561 - 128.3 + 570.2 - 117.81 = 885.19 ft²

Therefor, the most correct option is (B) 958.2 ft.².

4 0
3 years ago
PLEASE HURRY!!!!!!
Anni [7]

Answer:

Step-by-step explanation:

The manager visits a park and interviews people who are waiting to use the tennis courts.

7 0
3 years ago
A parabola has its focus at (1,2) and its directrix is y=-2. the equation of this parabola could be
Oksi-84 [34.3K]
<span>x^2/8 - x/4 + 1/8 = 0 A parabola is defined as the set of all points such that each point has the same distance from the focus and the directrix. Also the parabola's equation will be a quadratic equation of the form ax^2 + bx + c. So if we can determine 3 points on the parabola, we can use those points to calculate the desired equation. First, let's draw the shortest possible line from the focus to the directrix. The midpoint of that line will be a point on the desired parabola. Since the slope of the directrix is 0, the line will have the equation of x=1. This line segment will be from (1,2) to (1,-2) and the midpoint will be ((1+1)/2, (2 + -2)/2) = (2/2, 0/2) = (1,0). Now for the 2nd point, let's draw a line that's parallel to the directrix and passing through the focus. The equation of that line will be y=2. Any point on that line will have a distance of 4 from the directrix. So let's give it an x-coordinate value of (1+4) = 5. So another point for the parabola is (5,2). And finally, if we subtract 4 instead of adding 4 to the x coordinate, we can get a third point of 1-4 = -3. So that 3rd point is (-3,2). So we now have 3 points on the parabola. They are (1,0), (5,2), and (-3,2). Let's create some equations of the form ax^2 + bx + c = y and then substitute the known values into those equations. SO ax^2 + bx + c = y (1) a*1^2 + b*1 + c = 0 (2) a*5^2 + b*5 + c = 2 (3) a*(-3)^2 + b*(-3) + c = 2 Let's do the multiplication for those expressions. So (4) a + b + c = 0 (5) 25a + 5b + c = 2 (6) 9a - 3b + c = 2 Equations (5) and (6) above look interesting. Let's subtract (6) from (5). So 25a + 5b + c = 2 - 9a - 3b + c = 2 = 16a + 8b = 0 Now let's express a in terms of b. 16a + 8b = 0 16a = -8b a = -8b/16 (7) a = -b/2 Now let's substitute the value (-b/2) for a in expression (4) above. So a + b + c = 0 -b/2 + b + c = 0 And solve for c -b/2 + b + c = 0 b/2 + c = 0 (8) c = -b/2 So we know that a = -b/2 and c = -b/2. Let's substitute those values for a and c in equation (5) above and solve for b. 25a + 5b + c = 2 25(-b/2) + 5b - b/2 = 2 -25b/2 + 5b - b/2 = 2 2(-25b/2 + 5b - b/2) = 2*2 -25b + 10b - b = 4 -16b = 4 b = -4/16 b = -1/4 So we now know that b = -1/4. Using equations (7) and (8) above, let's calculate a and c. a = -b/2 = -(-1/4)/2 = 1/4 * 1/2 = 1/8 c = -b/2 = -(-1/4)/2 = 1/4 * 1/2 = 1/8 So both a and c are 1/8. So the equation for the parabola is x^2/8 - x/4 + 1/8 = 0 Let's test to make sure it works. First, let's use an x of 1. x^2/8 - x/4 + 1/8 = y 1^2/8 - 1/4 + 1/8 = y 1/8 - 1/4 + 1/8 = y 1/8 - 2/8 + 1/8 = y 0 = y And we get 0 as expected. Let's try x = 2 x^2/8 - x/4 + 1/8 = y 2^2/8 - 2/4 + 1/8 = y 4/8 - 1/2 + 1/8 = y 4/8 - 1/2 + 1/8 = y 1/2 - 1/2 + 1/8 = y 1/8 = y. Let's test if (2,1/8) is the same distance from both the focus and the directrix. The distance from the directrix is 1/8 - (-2) = 1/8 + 2 = 1/8 + 16/8 = 17/8 The distance from the focus is d = sqrt((2-1)^2 + (1/8-2)^2) d = sqrt(1^2 + -15/8^2) d = sqrt(1 + 225/64) d = sqrt(289/64) d = 17/8 And the distances match again. So we do have the correct equation of: x^2/8 - x/4 + 1/8 = 0</span>
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3 years ago
In expanded form, in kilometers, how far is Jupiter from the Sun?
DIA [1.3K]
<span>740,679,835 km i hope this helps you
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