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4 years ago

0.000006 write each number in scientific notation

1 answer:

4 0

Well, if you need the number (0.000006) to be more than 1 and less than ten, you need to move the decimal point. So, 0.000006-----> 6.0 and then, how places did you move the decimal to get 6.0 instead of 0.000006? Well, the answer is 6 so if you multiply 6.0*10 to the -6

The solution is 0.6*10 the the -6 power

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3 years ago

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What is the distance between -12 and 3

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15 units because I need to have

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3 years ago

Evaluate the following double integral: xy dA D where the region D is the triangular region whose vertices are (0, 0), (0, 3), (

natulia [17]

Answer:

I= 84

Step-by-step explanation:

for

$I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{}_{} \int\limits^{}_D {x*y} \, dx*dy$

since D is the rectangle such that 0<x<3 , 0<y<3

$I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{3}_{0} \int\limits^{3}_{0} {x*y} \, dx*dy = \int\limits^{3}_{0} {x} \, dx\int\limits^{3}_{0} {y} \, dy = x^{2} /2*y^{2} /2 = (3^{2} /2 - 0^{2} /2)* (3^{2} /2 - 0^{2} /2) = 3^{4} /4 = 81/4$

4 0

3 years ago

A ceiling light has a cross-section in the shape of a parabola. The parabola is 24 cm wide and 9 cm deep. The lightbulb is locat

Sedbober [7]

Answer:

4 cm

Step-by-step explanation:

The equation of a parabola with its vertex at the origin can be written as ...

y = 1/(4p)x^2

The problem statement tells us that one point on the parabola is (x, y) = (12, 9). We can put these values into the equation and solve for p, the distance from the focus to the vertex.

9 = 1/(4p)(12^2)

9×4/144 = 1/p = 1/4 . . . . . . . . multiply by the inverse of the coefficient of 1/p

Then p = 4, and the bulb is 4 cm from the vertex.

4 0

3 years ago

What whole number dimensions would allow the students to maximize the volume while keeping the surface area at most 160 square f

ycow [4]

Answer:

The whole number dimension that would allow the student to maximize the volume while keeping the surface area at most 160 square is 6 ft

Step-by-step explanation:

Here we are required find the size of the sides of a dunk tank (cube with open top) such that the surface area is ≤ 160 ft²

For maximum volume, the side length, s of the cube must all be equal ;

Therefore area of one side = s²

Number of sides in a cube with top open = 5 sides

Area of surface = 5 × s² = 180

Therefore s² = 180/5 = 36

s² = 36

s = √36 = 6 ft

Therefore, the whole number dimension that would allow the student to maximize the volume while keeping the surface area at most 160 square = 6 ft.

6 0

3 years ago