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Tamiku [17]
2 years ago
11

By how much does a a^3-2a exceed a^2 + a -6

Mathematics
1 answer:
olganol [36]2 years ago
8 0

Answer:

a³ - a² - 3a+ 6

Step-by-step explanation:

we are asked by how much does (a³-2a) exceed (a²+a-6)

in other words, we are being asked to find the difference between the 2 terms above.

Mathematically we are asked to find:

(a³-2a) - (a²+a-6)    (expand parentheses by distribution property)

=(a³-2a - a²- a + 6)

= a³ - a² - 3a+ 6    (answer)

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Find the limits of integration ly, uy, lx, ux, lz, uz (some of which will involve variables x,y,z) so that ∫uyly∫uzlz∫uxlxdxdzdy
motikmotik

Answer:

Hello your question is incomplete attached below is the complete question

Ix = 0   Ux = \sqrt{y-9z^2}

Iz = 0   Uz = \frac{\sqrt{y} }{3}

Iy = 5   Uy = 10

Step-by-step explanation:

Ix = 0   Ux = \sqrt{y-9z^2}

Iz = 0   Uz = \frac{\sqrt{y} }{3}

Iy = 5   Uy = 10

attached below is the detailed solution

4 0
2 years ago
injured runners train on a special track at a rehabilitation center. The track is a square with a half circle on its left and ri
diamong [38]

Answer:

The length of the track is approximately 51.7 ft

The track has <u>three</u> sides of the square and the distance round <u>a half of a</u> complete circle

Step-by-step explanation:

The given track shape and measurements are;

The shape on the left side of the track  = Square

The shape on the right side of the track  = Half circle

The area of the square on the the left side of the track  = 128 square feet

Therefore, from the area, A, of a square of side length, s, which is s × s, and letting the side length of the square = s, we have;

Area of the square portion of the track = s × s = s² = 128 ft²

Therefore, s = √(128 ft²) = 8·√(2) ft.

Whereby the side length of the square is bounded by the diameter of the half circle, we have;

Length of the diameter of the half circle = s = 8·√(2) ft.

The length of the perimeter of the half circle = π·D/2 = π × 8·√(2)/2 = π × 4·√(2) ≈ 17.77 ft.

The perimeter of the track, which is the length of the track is made up of the three sides of the square opposite to the half circle and the circumference of the half circle.

Therefore;

The length of the track = 3 × 8·√(2) ft + π × 4·√(2) ft. = 4·√2×(π+6) ≈ 51.7 ft

The length of the track ≈ 51.7 ft

Which gives;

The track has <u>three</u> sides of the square and the distance round <u>a half of a</u> complete circle.

5 0
2 years ago
Uri shared 6/8 of his orange with a friend. uri ste the rest so how much of the orsnge did he eat
Klio2033 [76]
He ate 1 - 6/8 which is 2/8, or 1/4 of an orange.
6 0
2 years ago
Read 2 more answers
To pass inspection, a machine part from a factory must be no thinner than 4.225 centimeter and no thicker than 4.233 centimeters
Mrrafil [7]
Machine should abide by te following:


4.225<Machine<4.233  The margin of acceptance
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4.233-4.225 = 0.008cm.


Any thickness greater than 4.233+0.008 or smaller than 4.225-0.008

or <4.233cm  & >4.217
6 0
2 years ago
Round 720,051,852 to the nearest hundred
marysya [2.9K]

Answer:

720,051,900

Step-by-step explanation:

it rounds up to 720,051,900 because 5 tells you to round up

4 0
1 year ago
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