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4 years ago

One number is equal to the square of another. Find the numbers if both are positive and their sum is 380.

Mathematics

1 answer:

Paladinen [302]4 years ago

7 0

Answer:

19 and -20

Step-by-step explanation:

Call that number is x, we have:

$x + x^2 = 380\\(x^2 + 2*x*\frac{1}{2} + \frac{1}{4} ) -\frac{1521}{4} =0\\(x+\frac{1}{2})^2 =\frac{1521}{4} \\First situation: x + \frac{1}{2} = \sqrt{\frac{1521}{4} } =\frac{39}{2} \\x = \frac{39}{2} - \frac{1}{2} = 19\\Second situation: x + \frac{1}{2} =- \sqrt{\frac{1521}{4} } =-\frac{39}{2} \\\\x =- \frac{39}{2} - \frac{1}{2} = -20\\\\In conclusion, x= 19 and x=-20$ Brainliest, please? It costs me so much time!

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3 years ago

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3 years ago

Please Helppp Cacl AB Determining Slopes.

MrMuchimi

The zeros of the function are those values of $x$ that make $y=0$ . So we solve

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The denominator will always be positive, so we can multiply both sides of the equation by it to get

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where $n$ is any integer. If we take $n=\pm1$ we should get the two solutions immediately adjacent to the one near $x=0$ that still lie in the interval $-5\le x\le5$ . So the other two zeros are $x=-\arcsin\dfrac1{50}\pm\pi$ .

The tangent line to the curve at any $x$ is determined by the value of the derivative of the function at that value of $x$ . So first compute the derivative:

$y=\dfrac{1+50\sin x}{x^2+3}\implies y'=\dfrac{50\cos x(x^2+3)-2x(1+50\sin x)}{(x^2+3)^2}=\dfrac{(50x^2+150)\cos x-100x\sin x-2x}{(x^2+3)^2}$

Now just plug in the values of $x$ determined above. It's helpful to note

$\cos\left(\arcsin\dfrac1{50}\right)=\dfrac{7\sqrt{51}}{50}$
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3 years ago