3 years ago

I need help answer ASAP I’m confused

Mathematics

1 answer:

katrin [286]3 years ago

4 0

Answer:

Step-by-step explanation:

You can plug in the 294 for A and solve

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tino4ka555 [31]

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2 years ago

Help please urgent 25 point

uranmaximum [27]

Answer:

Step-by-step explanation:

b = the hypotenuse of a right angle triangle

a = the adjacent side.

R is being defined by the cosine

cos(R) = adjacent / hypotenuse

adjacent = 16* sqrt(2)

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2 years ago

Solve the initial value problem: y'(x)=(4y(x)+25)^(1/2) ,y(1)=6. you can't really tell, but the '1/2' is the exponent

goblinko [34]

Answer:

$y(x)=x^2+5x$

Step-by-step explanation:

Given: $y'=\sqrt{4y+25}$

Initial value: y(1)=6

Let $y'=\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\sqrt{4y+25}$

Variable separable

$\dfrac{dy}{\sqrt{4y+25}}=dx$

Integrate both sides

$\int \dfrac{dy}{\sqrt{4y+25}}=\int dx$

$\sqrt{4y+25}=2x+C$

Initial condition, y(1)=6

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Put C into equation

Solution:

$\sqrt{4y+25}=2x+5$

$4y+25=(2x+5)^2$

$y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}$

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Hence, The solution is $y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}$ or $y(x)=x^2+5x$

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3 years ago

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2 years ago

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