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Anton [14]
3 years ago
8

Round 59.983 to the nearest tenth

Mathematics
1 answer:
Margarita [4]3 years ago
6 0

Answer:60.0

Step-by-step explanation:

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The tartness of a tangerine
MAVERICK [17]
What do you want us to figure out?
3 0
4 years ago
The domain of f(x) is the set of all real values except 7, and the domain of g(x) is the set of all real values except -3. Which
Savatey [412]

Answer:

Third choice

Step-by-step explanation:

The domain of f(x) is the set of all real values except 7, i.e x\ne7

and the domain of g(x) is the set of all real values except -3, i.e x\ne -3

The domain of g\circ f is the set of all real numbers x in the domain of f  such that f(x) is in the domain of g(x).

However any numbers that are excluded from the domain of f must also be excluded from the domain of g\circ f.

Therefore the correct choice is all real values except x=7 and the x for which f(x) =7

6 0
4 years ago
Read 2 more answers
2x+2y=-2 y = x + 5 please hurry!
larisa [96]

Answer:

2x+2y=x+5

(move the x next to the 2y)

= 2x+2y-x=5

(the x is negative because i moved it)

(collect the like terms)

= x+2y=5

(so the answer is...)

x+2y=5

6 0
3 years ago
Write 7 x 10^-3<br> in standard notation.
VashaNatasha [74]

Answer:

0.007

Step-by-step explanation:

A negative notation starts with 0's from the left.

count 3 places before the 7, that is where my point is.

I am pretty sure of my explanation but the answer is correct.

5 0
3 years ago
find the volume of a solid that is generated by rotating around the indicated axis the plane region bounded by the giveb curves:
Fofino [41]
Assuming the area below the line y=0 (i.e. x>1) does NOT count, the area to be rotated is shown in the graph attached.

A. Again, using Pappus's theorem,
Area, A = (2/3)*1*(1-(-1))=4/3  (2/3 of the enclosing rectangle, or you can integrate)
Distance of centroid from axis of rotation, R = (2-0) = 2
Volume = 2 π RA = 2 π 2 * 4/3 = 16 π / 3 (approximately = 16.76 units)

B. By integration, using the washer method
Volume = 2\pi\int_{-1}^1(1-x^2)(2-x)dx
=2\pi\int_{-1}^1(x^3-2x^2-x+2)dx
=2\pi[x^4/4-2x^3/3-x^2/2+2x]_{-1}^{1}
=2\pi([1/4-2/3-1/2+2]-[1/4+2/3-1/2-2])
=2\pi(8/3)
= 16 π /3   as before



3 0
3 years ago
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