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Anvisha [2.4K]
3 years ago
14

The system of equations shown below is graphed on a coordinate grid:

Mathematics
2 answers:
Vlad1618 [11]3 years ago
8 0

Answer:

um idek

Step-by-step explanation:

DaniilM [7]3 years ago
5 0
4y + x = 4 . . . . . . . (1)
2y - x = 8 . . . . . . . (2)

(1) + (2) => 6y = 12
y = 12/6 = 2

From (2), 2(2) - x = 8
4 - x = 8
x = 4 - 8 = -4

Therefore solution is (-4, 2) and lies on both lines.
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How to solve this inequality: 3p - 6 > 21.
bearhunter [10]
First of all you do +6 to both sides which will get you now 3p > 27 then you do divide by 3 to get the p by itself, which will give you answer of p > 9, LETTER C
5 0
3 years ago
Please help me i really need help
Aleks [24]

Answer:

The answere is C

Step-by-step explanation:

6 0
3 years ago
Please help me with this question, thanks.
arlik [135]

Answer:

  2.48×10^13 miles

Step-by-step explanation:

There are about (365.25 days)(86400 seconds/day) = 31,557,600 seconds in one year. There are 1609.344 meters in one mile. So, the conversion can be written as ...

  (4.22 ly) (3×10^8 m/s) (3.15576×10^7 s/yr) / (1.609344×10^3 m/mi)

  = 4.22×3×3.15576/1.609344×10^(8+7-3) mi

  ≈ 24.8×10^12 mi

4.22 light years is about 2.48×10^13 miles

8 0
3 years ago
The expression 2x2 + 12x + 10 represents a bird's height, in feet, x seconds after it leaves its nest,
marusya05 [52]
The answer is A. You add the extra 10 feet to however high the bird flew from there
8 0
3 years ago
Find the length of the arc of the circular helix with vector equation r(t) = 2 cos t i + 2 sin t j + tk from the point (2, 0, 0)
USPshnik [31]

\vec r(t)=2\cos t\,\vec\imath+2\sin t\,\vec\jmath+t\,\vec k

\implies\vec r'(t)=-2\sin t\,\vec\imath+2\cos t\,\vec\jmath+\vec k

\implies\|\vec r'(t)\|=\sqrt{(-2\sin t)^2+(2\cos t)^2+1^2}=\sqrt5

Then the length of the arc is

\displaystyle\int_0^{2\pi}\|\vec r'(t)\|\,\mathrm dt=\sqrt5\int_0^{2\pi}\mathrm dt=2\sqrt5\,\pi

4 0
3 years ago
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