Advanced Placement (AP), 11.11.2019 22:31 amayarayne5

Hello i am a 15 year old i am wondering if i should tell mom about a girl i really like and want to start dating or if i should avoid bringing it up because she doesn't want me to date till i am 19 and i know the girl likes me also what should i do i have known the girl for 3 years and have fealt this way for 2 years

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Answer from: gavianacandelar8522

Wait until you think it is the best time to ask. I would think when you are 18-19.

Answer from: isisg8

Explanation:

If you like her and you think she likes you go ahead and start dating. You might find out you had less in common than you thought so if it doesn't work out no harm done. And after a few weeks if things seem to be moving smoothly go ahead and tell your parents.

Answer from: Quest

1. unlikely2. unlikely

Answer from: Quest

Answer: the interval on which the curve is concave up is (-∞, -1/2)====== $\displaystyle y = \int_{0}^x \frac{1}{1 + t + t^2} \; dt$ we find the interval where this curve is concave up by finding the interval in which the 2nd derivative is positive.differentiate the curve equation with respect to x $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \end{aligned}$ let the function f denote be the anti-derivative of the inside expression, i.e. a function such that $(d/dx)f(x) = 1/(1 + t + t^2)$ .then use the fundamental theorem of calculus: $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \\& = \frac{d}{dx} \big[ f(x) - f(0) \big] & & \text{\footnotesize apply ftc} \\ & = \frac{d}{dx}f(x) - 0 \\ & = \frac{1}{x + x + x^2} & & \text{\footnotesize since $\frac{d}{dx}f(x) = \frac{1}{1 + x + x^2}$ } \end{aligned}$ differentiate again to get second derivative. we can use chain rule with 1/x as the outside function and 1 + x + x² as the inside function $\begin{aligned} \frac{d}{dx} \left( \frac{dy}{dx} \right) & = \frac{d}{dx}\left(\frac{1}{1 + x + x^2} \right) \\ \frac{d^2y}{dx^2} & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot \tfrac{d}{dx}(1 + x + x^2) \\ & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot (1 + 2x) \\ & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \end{aligned}$ find critical numbers of the second derivative. these numbers are where the 2nd deriv. is undefined or zero.where the 2nd derivative is zero: set (d²y/dx²) to equal 0 and solve for x $\begin{aligned} 0 & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \\ 0 & = -(1+2x) \\ -1 & = 2x \\ x & = -1/2 \end{aligned}$ where the 2nd derivative is undefined: if the denominator is equal to zero, the derivative will be undefeined. attempting to solve the equation 1 + x + x² = 0results in a discriminant of $b^2 - 4ac = 1^2 - 4(1)(1) = -3$ meaning that the equal 1 + x + x² = 0 has no real solutions and therefore will never equal zero.======our only critical number is x = -1/2.this forms two intervals to test the sign for: (-∞, -1/2) and (-1/2, ∞)test values into the 2nd derivative for (-∞, -1/2)if we try x = -1, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = -1} & = \frac{-(1+2(-1))}{\left(1+(-1)+(-1)^2\right)^2} \\ & = \frac{-(1-2)}{1} \\ & = 1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave up on (-∞, -1/2)test values into the 2nd derivative for (-1/2, ∞)if we try x = 0, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = 0} & = \frac{-(1+2(0))}{\left(1+(0)+(0)^2\right)^2} \\ & = \frac{-(1)}{1} \\ & = -1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave down on (-∞, -1/2).the interval on which the curve is concave up is (-∞, -1/2)

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