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3 years ago

Select the correct answer. What is the definition of csc θ ?

Mathematics

1 answer:

OleMash [197]3 years ago

8 0

Answer:

A. csc θ=1/sinθ

Step-by-step explanation:

From SOHCAHTOA,

sine = opposite/ hypotenuse

Cos = adjacent / hypotenuse

Tan = opposite / adjacent

In a right angled triangle, the cosecant of an angle is the length of the hypotenuse divided by the length of the side opposite the angle. This is the inverse of sine , therefore:

Csc Ф = 1/ sinФ

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Elements in group 17 on the period table will most likely bond with elements in group

andreev551 [17]

Answer:Elements in group 17 on the period table will most likely bond with elements in group 1.

Step-by-step explanation:

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3 years ago

What is the range of the function y=x^2

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Domain: (-∞, ∞) or {X, where x ∈ R}
Range: (-∞, ∞) or {Y, where y ∈ R}

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3 years ago

Read 2 more answers

Click screenshot for the problem answer 13

Vedmedyk [2.9K]

24P÷30
24×12÷30
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288÷30=9.6
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3 years ago

The marked price of an article is Rs.2080. After allowing d% discount and levying

mash [69]

Answer:

13%

Step-by-step explanation:

Given:

MP = 2080
Discount = d%
VAT = (d-2)%
Cost = 1997.84

Apply discount:

2080 - d% = 2080*(1 - 0.01d)

Add VAT:

2080*(1 - 0.01d) + (d - 2)%
2080*(1 - 0.01d) * (1 + (d -2)/100)
2080*(1 - 0.01d) * (0.98 + 0.01d) = 1997.84
(1 - 0.01d)(0.98 + 0.01d) = 1997.84/2080
0.98 + 0.01d - 0.0098d - 0.0001d² = 0.9605
- 0.0001d² + 0.0002d + 0.98- 0.9605 = 0
0.0001d²- 0.0002d - 0.0195 = 0
d² - 2d + 195 = 0

Solving the quadratic equation we get:

d = 15

Then

VAT rate = 15 - 2 = 13%

7 0

3 years ago

Find dy/dx for 4 - xy = y^3

storchak [24]

Answer:

$\frac{dy}{dx}=-\frac{y}{3y^2+x}$

Step-by-step explanation:

4-xy=y^3

dy/dx=?

$\frac{d(4-xy)}{dx}=\frac{d(y^3)}{dx}\\ \frac{d(4)}{dx}-\frac{d(xy)}{dx}=3y^{3-1}\frac{dy}{dx}\\ 0-(\frac{dx}{dx}y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(1y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -y-x\frac{dy}{dx}=3y^2\frac{dy}{dx}$

Solving for dy/dx: Addind x dy/dx both sides of the equation:

$-y-x\frac{dy}{dx}+x\frac{dy}{dx}=3y^2\frac{dy}{dx}+x\frac{dy}{dx} \\ -y=3y^2\frac{dy}{dx}+x\frac{dy}{dx}$

Common factor dy/dx on the right side of the equation:

$-y=(3y^2+x)\frac{dy}{dx}$

Dividing both sides of the equation by 3y^2+x:

$\frac{-y}{3y^2+x}=\frac{(3y^2+x)}{3y^2+x}\frac{dy}{dx}\\ -\frac{y}{3y^2+x}=\frac{dy}{dx}\\ \frac{dy}{dx}=-\frac{y}{3y^2+x}$

7 0

3 years ago