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3 years ago

PLEASE HELP im so lost

Mathematics

2 answers:

Nana76 [90]3 years ago

8 0

Answer:

y=1x-5

Step-by-step explanation:

the y-intercept is -5 and it goes up one square unit every time

Likurg_2 [28]3 years ago

3 0

Answer:

y = x + (-5)

Step-by-step explanation:

I can math.

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2 years ago

How does the formula S=2B+Ph apply to finding the surface area of a right prism?

tigry1 [53]

Answer S−2BP=h

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2 years ago

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.
$(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}$
$\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}$
$2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C$

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

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

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

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

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

$\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}$

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3 years ago

A rectangular floor space in a community center is 40 ft long and 32 ft wide. A volunteer is setting up chairs on the floor spac

pav-90 [236]

It is B because 40x32=1280 and 5x4=20 then divide 1280 by 20 and you get 64 but you can’t have 64 so you would need to only have 60 so the answer is B

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3 years ago

How much is 11+5+3x2+8x1?

adoni [48]

11+5=16 3x2= 6 8x1= 8 so add it together, 16 + 6 + 8 = 30.

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3 years ago

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