All categories

3 years ago

What is the slope of a line y=7.8x+62

Mathematics

1 answer:

dexar [7]3 years ago

3 0

Answer:

m = 7.8

Step-by-step explanation:

Its the slope. Its the number next to the x.

You might be interested in

Consider 6x2 + 6x + 1. Which term immediately tells you that this expression is NOT a perfect square trinomial? Justify your ans

oksian1 [2.3K]

Answer:

Step-by-step explanation:

The 6x^2 because 6 is not a perfect square.

4 0

3 years ago

Read 2 more answers

At a bike shop, Kirby is buying a new bike that costs

mart [117]

Answer:

tax: $18.9

total cost: $333.9

Step-by-step explanation:

$315 × 0.06 = 18.9

18.9 + 315 = 333.9

btw you can also do "315 × 1.06" to get the total cost without having to manually add the tax

3 0

3 years ago

2 3/4 x 8 = blank/4 x 8/1

Alex

Answer: 11

Step-by-step explanation:

2 3/4 × 8 = p/4 × 8/1

Let p be the missing number

Convert the mixed fraction 2 3/4 to proper fraction, multiply 4 x 2, then add 3= 11/4

11/4 × 8 = p/4 ×8

p= 11

I hope this helps.

5 0

3 years ago

Read 2 more answers

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

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

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

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