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Fofino [41]
3 years ago
9

What is oneplus two?​

Mathematics
2 answers:
alexdok [17]3 years ago
8 0

Answer:

one plus two is three

Step-by-step explanation:

1+2 = 3

nlexa [21]3 years ago
6 0

Answer: 3

Step-by-step explanation: take the numbers 1 (one) and 2 (two) then add them together, we know that’s 1 + 1 = 2 so If we do 1 + 2 or 1 + 1 + 1 we’ll get 3 as the final answer

Hope this helps, a marking as brainliest and / or a star rating would be appreciated! :)

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Complete the pattern using powers of 5
krok68 [10]
1/5^2 = 1/25
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3 years ago
Solve the following equation. 3x/7-2=15 (Must be a mixed number)
Elina [12.6K]

Answer:

Mixed Number

39 \frac{2}{3}

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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
Show whether or not y=x+3 is tangential to the curve y^2=x​
wolverine [178]

The line y = x + 3 has slope 1, so we look for points on the curve where the tangent line, whose slope is dy/dx, is equal to 1.

y² = x

Take the derivative of both sides with respect to x, assuming y = y(x) :

2y dy/dx = 1

dy/dx = 1/(2y)

Solve for y when dy/dx = 1 :

1 = 1/(2y)

2y = 1

y = 1/2

When y = 1/2, we have x = y² = (1/2)² = 1/4. However, for the given line, when y = 1/2, we have x = y - 3 = 1/2 - 3 = -5/2.

This means the line y = x + 3 is not a tangent to the curve y² = x. In fact, the line never even touches y² = x :

x = y²   ⇒   y = y² + 3   ⇒   y² - y + 3 = 0

has no real solution for y.

3 0
2 years ago
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