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3 years ago

I need the steps to this question.

Mathematics

1 answer:

yanalaym [24]3 years ago

3 0

im not sure but maybe look up how to calculate volume of a cone.. then go from there :D

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Find the value of X.

Vera_Pavlovna [14]

Answer:

x=16

Step-by-step explanation:

(4x+2)+(x+8)=90

5x+10=90

5x=80

x=16

3 0

3 years ago

3. What is the equation in slope-intercept form of the line that passes through the

zavuch27 [327]

The Answer: y=1/4x+3

3 0

2 years ago

How do i solve this? Please give me step by step explanation.

Viktor [21]

Answer:1.5

Step-by-step explanation:

12.6j + 8 = 10.2j +4.4

2.4j=3.6

j=3.6/2.4= 1.5

3 0

2 years ago

Read 2 more answers

a jacket originally sold for $45. this week it went on sale for 20% off. what is the discount and what is the sales price?

Strike441 [17]

Answer:

discount = 9

new price = 36

Step-by-step explanation:

The discount is the price times the discount percent

45 * 20%

Change to decimal form

45*.20

The new price is the original price minus the discount

45-9 = 36

4 0

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

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

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

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

6 0

3 years ago