Advanced Placement (AP), 31.01.2020 05:54, bamagirl3871

When setting up a query, setting "criteria" is the same as setting a select one: a. data type that will be saved with the query. b. sort that will be saved with the query. c. filter that will be saved with the query. d. table that will be saved with the query. question 12 not yet answered marked out of 1.0 not flaggedflag question question text in a query, the sort order of multiple fields is determined by select one: a. the order in which the fields appear from left to right in query design view. b. the order in which the fields appear from left to right in datasheet view. c. the order in which the fields appear from right to left in query design view. d. you cannot sort by more than one column. question 13 not yet answered marked out of 1.0 not flaggedflag question question text how do you create new simple query in ms access? select one: a. go to the navigation pane and double-click on the query title. b. click on the little arrow at the top of the data field. c. go to the "file" tab and click "new". d. go to the "create" tab and choose "query design". question 14 not yet answered marked out of 1.0 not flaggedflag question question text a query can pull its information from select one: a. one or more queries in the database. b. one or more queries and/or tables in the database. c. only a single table in the database. d. one or more tables in the database. question 15 not yet answered marked out of 1.0 not flaggedflag question question text you have created a new query and saved it as customerq. you then add a specific criteria and want to save the new query as customernewq. how do you save the new query without overwriting the old query? select one: a. go to "file", then click "save object as", then give the query a new name. b. click the blue floppy disk "save" button at the top left. c. go to "file", then click "save as", then give the query a new name. d. go to "file", then click "save", then give the query a new name.

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Iam guessing it’s a metallic because it is powdery and so belongs to group 2it is also light. so i would guess it’s graphite.if available, additional information would be very for-example its color. if black it would be division e. also try to mark it on a paper.

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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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Ibelieve the answer would be a

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When setting up a query, setting "criteria" is the same as setting a select one: a. data type that...