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Naddik [55]
3 years ago
15

What is the value of z in the equation 2z + 6 = –4? (5 points) 5 1 –1 –5

Mathematics
2 answers:
PtichkaEL [24]3 years ago
6 0
The answer is -5. Hope this helps.
marysya [2.9K]3 years ago
6 0
Plug in the options
not 5( 2 x 5) + 6= 16
not 1 ( 2 x 1) + 6 = 8
not -1 (2 x -1) + 6= 4
so its 5 ( 2 x -5) + 6= -4
- 
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Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck
Nimfa-mama [501]
\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}
\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}

Now, isolate the variables, so you can integrate.
(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}
\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}
2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C


4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C
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y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}
y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}

At x = 1, y = 0.
0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}
-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}

4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0
\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}


4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C
0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C
C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}
C = -\frac{\pi}{2}

\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}
6 0
3 years ago
Please help, and show work if you can!! :))
dmitriy555 [2]

Answer:

Step-by-step explanation:

ΔABD ≅ ΔCBD .

So AD = DC

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     8x = 3x + 5 {subtract 3s from both sides}

  8x - 3x= 5

        5x = 5 {divide both sides by 5}

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x = 1

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