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3 years ago

What is the value of x in the equation 0, when y = 15? 4.

Mathematics

1 answer:

Tatiana [17]3 years ago

8 0

Answer:

x =200

Step-by-step explanation:

$\frac{1}{5}x - \frac{2}{3}(15) =30\\\\\frac{1}{5}x-10=30\\\\\frac{1}{5}x=30+10\\\\\frac{1}{5}x=40\\\\x = 40*5\\x= 200$

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Answer:

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Step-by-step explanation:

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dezoksy [38]

Answer:

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3 years ago

Read 2 more answers

Find dy/dx if y =x^3+5x+2/x²-1

stiks02 [169]

<u>Differentiate using the Quotient Rule</u> –

$\qquad$ $\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\$

According to the given question, we have –

f(x) = x^3+5x+2
g(x) = x^2-1

Let's solve it!

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2) \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5) - ( x^3+5x+2) 2x}{(x^2-1)^2 }\\$

$\qquad$ $\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\$

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\$

$\qquad$ $\pink{\therefore \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}} = \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\$

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2 years ago

Find the values of x and y

icang [17]

Answer:

$x=15\\y=38$

Step-by-step explanation:

By definition, the sum of the interior angles of a triangle is 180 degrees.

You can observe that the given triangle has an angle that measures 90 degrees and it also has two equal sides.This means that the other two angles are equal.

Knowing this, you can conclude that the triangle is an Isosceles Right Triangle and the other angles measure 45 degrees each.

Therefore, now you know that:

$3x=45$