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Lelu [443]
3 years ago
6

Can someone help me with this one question

Mathematics
2 answers:
Ludmilka [50]3 years ago
8 0
Idk hahah but an i dont know that shift
erma4kov [3.2K]3 years ago
4 0
Second option is the correct one
a1 = 37 \\ an = a(n - 1) - 11
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I am doing online class and I am looking for a little help please
kati45 [8]
The domain is about how far left-to-right the graph goes.
In relation to the x-axis, the graph starts at x = –3 (with an open circle at –3) and then continues over to the right forever.
This is the shown in the picture with the red markup.
In interval notation, this is (-3, infinity).

Remember to use that left-to-right orientation for interval notation!

The range is in turn about how low to how high the graph goes.
On the graph, I’d do the same thing I did on the red marked up graph and compare the graph to the y-axis.
The graph starts down at y = –5 (with an open circle at –5) and then continues on up forever.
In interval notation, this is (-5, infinity).

4 0
2 years ago
7×300=7× blank hundreds
VladimirAG [237]
7x300=7x 3 hundres because in 300 hundreds there is 3 hundreds.I know that didnt make sense.so basically just look at the front number number its 3 so 3 hundreds but if your working with thousands and the front number is 5 its 5 thousands.I hope this was helpful.
6 0
3 years ago
Which symbol replaces the box to make the statement true? 2⋅18−18□2⋅[28−3⋅(2+4)] <br>A. <br>C. =
Misha Larkins [42]

Answer:

2⋅18−18 is less than 2⋅[28−3⋅(2+4)]

Step-by-step explanation:

7 0
3 years ago
The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati
makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\

And the second one would be

y = 2-x \,\,\,\,\,,  1 \leq x \leq 2.

If you rotate the first region around the "y" axis you get that

{\displaystyle A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51 }

And if you rotate the second region around the "y" axis you get that

{\displaystyle A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188 }

And the sum would be  2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first  region around the x axis you get that

{\displaystyle A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708 }

And if you rotate the second region around the x axis you get that

{\displaystyle A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472 }

And the sum would be 1.5708+1.0472 = 2.618

7 0
3 years ago
92.3-(3.2 divided by 0.4) times 8
sp2606 [1]

Answer:

28.3 that's the answer

Step-by-step explanation:

3 0
3 years ago
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