4. In the figure below, the measure of angle A is 35°. If the

1 answer:

5 0

WILL BRAINLIEST! psychology...
a description of the three topics, concepts, or theories you learned about in the module that you consider most important; choose topics, concepts or theories from at least three lessons in the module
an explanation of how what you have learned about these three topics, concepts, or theories can be applied in your own life or in the lives of others close to you
at least one paragraph for each topic, concept, or theory, for a total of at least three paragraphs; use this organizer to help you format your work

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Slope integrals for calculus, having trouble need all the help I can get

Harlamova29_29 [7]

(1) The integral is straightforward; x ranges between two constants, and y ranges between two functions of x that don't intersect.

$\displaystyle\int_{-2}^1\int_{-x}^{x^2+2}\mathrm dy\,\mathrm dx$

(2) First find where the two curves intersect:

y ² - 4 = -3y

y ² + 3y - 4 = 0

(y + 4) (y - 1) = 0

y = -4, y = 1 → x = 12, x = -3

That is, they intersect at the points (-3, 1) and (12, -4). Since x ranges between two explicit functions of y, you can capture the area with one integral if you integrate with respect to x first:

$\displaystyle\int_{-4}^1\int_{y^2-4}^{-3y}\mathrm dx\,\mathrm dy$

(3) No special tricks here, x is again bounded between two constants and y between two explicit functions of x.

$\displaystyle\int_1^5\int_0^{\frac1{x^2}}\mathrm dy\,\mathrm dx$

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3 years ago

Grandmother poem Pg 27 Read

goldfiish [28.3K]

Answer:

I would feel sad that my friends didn't have time for me. They would feel like i did care about them. about 12 female young

Explanation:

this is what I think

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3 years ago

Particle x moves along the positive x-axis so that its position at time t ~ 0 is given by x(t) = 5t 3 - 9t 2 + 7. at what time t

Arada [10]

Answer: 1.2 seconds

Explanation:

$At the farthest point the velocity of particle will be zero. \ Differentiating the given equation of position and equating it with zero:$$\begin{aligned}x(t) &=5 t^{3}-9 t^{2}+7 \\x^{\prime}(t) &=15 t^{2}-18 t \\0 &=15 t^{2}-18 t \\t(15 t-18) &=0 \\t &=0 \\t &=\frac{18}{15}=1.2 \mathrm{~s}\end{aligned}$

$$$Thus, at $t=1.2 \mathrm{~s}$ particle will be at the farthest point. \\\\\underline{Justification:}\\The particle moves towards the left from $t \in[0,1.2)$. After that particle keeps moving in the positive x-direction.$