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3 years ago

The number of basic trigonometric ratios is.... A.3 B.4 C.5 D.6

Mathematics

2 answers:

jasenka [17]3 years ago

8 0

There’s three
Sin, cousine, and tangent
:)
Mark me brainliest please :/

Eddi Din [679]3 years ago

3 0

Answer:

There are three basic trigonometric ratios: sine , cosine , and tangent .

Step-by-step explanation:

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Answer:

Step-by-step explanation:

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2 years ago

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ArbitrLikvidat [17]

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3 years ago

Read 2 more answers

Find the general solution of the following equation. Express the solution explicitly as a function of the independent variable.

Mashcka [7]

Answer:

Therefore the general solution is

$2 \sqrt w = 2 ln(x) - 5 \frac1x +c$

Step-by-step explanation:

Integration Rule:

$\int x^n dx= \frac{x^{n+1}}{n+1}+c$
$\int \frac1x dx= ln(x) +c$

Given differential equation is

$x^2 \frac{dw}{dx}= \sqrt{w}(2x+5)$

$\Rightarrow x^2 dw= \sqrt{w} (2x+5) dx$ [ multiplying dx both sides]

$\Rightarrow \frac{dw}{\sqrt w}= \frac{(2x+5)}{x^2} dx$ [ dividing $x^2\sqrt w$ both sides]

Integrating both sides

$\int \frac{dw}{\sqrt w}=\int \frac{(2x+5)}{x^2} dx$

$\Rightarrow \int w^{-\frac12} dw=\int (\frac{2x}{x^2}+\frac{5}{x^2} )dx$

$\Rightarrow \int w^{-\frac12} dw=\int \frac{2}{x}dx +\int\frac{5}{x^2} dx$

$\Rightarrow \frac{w^{-\frac12+1}}{-\frac12+1} =2ln x+5 \frac{x^{-2+1}}{-2+1}+c$ [ c is arbitrary constant]

$\Rightarrow 2 \sqrt w = 2 ln(x) - 5 \frac1x +c$

Therefore the general solution is

$2 \sqrt w = 2 ln(x) - 5 \frac1x +c$

6 0

3 years ago

Help me please, I beg of you. ;w;

Soloha48 [4]

The solution should be x=-1 y=-2

8 0

2 years ago

Please answer these questions! Explanations are really helpful but not needed (: Doing one or even a couple problems, maybe all

Lisa [10]

Answer:

which one is the question I can't understand

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2 years ago