Advanced Placement (AP), 25.06.2019 22:00 naruto63

Which part of each sentence is a dependent clause? the boys didn’t think dani could win the race even though she was on the school track team. trisha went down to the farmer’s market so she could get some apples. after he was injured by a stray ball, alex was sent to the nurse’s office. though he hated math, chris tried to do the equations.

Answers: 1

Show answers

Answers

The dependent clause of a sentence is a part of a sentence before or after a comma that can't stand alone. the dependent clauses are: " After he was injured by a stray ball," and "Though he hated Math,"

Answer from: homework2358

'even though she''so she could'after he was ''though he hated math;''

Explanation:

edmentum

Answer from: Quest

ek = (m.v2) / 2

v - velocity of system after the collision

m = m1 + m2

ek = (m1 + m2) v2/2

ek = (200 + 250) 2,32/2 = 450*5,29/2 = 1190,25 j

by: nicholas dooley

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)

Another question on Advanced Placement (AP)

Advanced Placement (AP), 22.06.2019 11:10

Atwo point turn is similar to a three point turn save for which of the following aspects

Answers: 1

Answer

Advanced Placement (AP), 22.06.2019 23:50

Ayellow line means traffic on either side may cross the center line to pass another vehicle. a. single solid b. double solid c. single broken d. double broken

Answers: 2

Answer

Advanced Placement (AP), 23.06.2019 05:20

On a certain portion of an experiment, a statistical test result yielded a p-value of 0.18. what can you conclude? 2(0.18) = 0.36 < 0.5; the test is not statistically significant. if the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 18% of the time, so the test is not statistically significant. if the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 82% of the time, so the test is not statistically significant. 0.18 > 0.05; the test is statistically significant. p = 1 - 0.18 = 0.82 > 0.05; the test is statistically significant.

Answers: 2

Answer

Advanced Placement (AP), 24.06.2019 06:00

In a government based on the consent of the governed, who has the ultimate authority?

Answers: 2

Answer

You know the right answer?

Which part of each sentence is a dependent clause? the boys didn’t think dani could win the race ev...

Questions