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3 years ago

Can someone please answer. There is one problem. There's a picture. Thank you!

Mathematics

1 answer:

IrinaK [193]3 years ago

3 0

$\tan x= 0 \Rightarrow x=k\pi,k\in\mathbb{Z}\\\\
\pi \text{ rad}=180^{\circ}$
so zeroes of tangent function are integer multiples of 180.
Therefore the correct answer is the last one.

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3 years ago

Yall can u please help me this is a grade i really need All answers before 9:30

il63 [147K]

Answer:

Step-by-step explanation:

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2 years ago

PLEASE HELP ME!

kotegsom [21]

Answer:

Bottom left graph

Step-by-step explanation:

We have to use what is called the zero-interval test [test point] in order to figure out which portion of the graph these inequalities share:

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+ 2x + 2x

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−2[0] + 0 ≤ 4

0 ≤ 4 ☑ [We shade the part of the graph that CONTAINS THE ORIGIN, which is the right side.]

$\displaystyle y > x + 2 \\ \\ 0 ≯ 2$ [We shade the part of the graph that does not contain the origin, which is the left side.]

So, now that we got that all cleared up, we can tell that the graphs share a region in between each other and that they both have POSITIVE RATE OF CHANGES [SLOPES], therefore the bottom left graph matches what we want.

** By the way, you meant $\displaystyle y > x + 2$ because this inequality in each graph is a dashed line. It is ALWAYS significant that you be very cautious about which inequalities to choose when graphing. Inequalities can really trip some people up, so once again, please be very careful.

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3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. Its height

Lina20 [59]

Answer:

Instantaneous Velocity at t = 1 is 20 feet per second

Step-by-step explanation:

We are given he following information in the question:

$y(t)=43t-23t^{2}$

B) Instantaneous Velocity at t = 1

$y(1) = 43(1)-23(1)^2 = 20$

A) Formula:

Average velocity =

$\displaystyle\frac{\text{Displacement}}{\text{Time}}$

1) 0.01

$y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}$

2) 0.005 s

$y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}$

3) 0.002 s

$y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}$

4) 0.001 s

$y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}$

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2 years ago

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Elena L [17]

Answer:

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Step-by-step explanation:

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2 years ago