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3 years ago

Which statements are true about the linear inequality y>

Mathematics

2 answers:

Nikitich [7]3 years ago

6 0

For this case we have the following inequality:

$y> \frac {3} {4} x-2$

It is observed that inequality does not include equality, so we will have a dotted line at the border of the graph.

The slope of the line is given by $m = \frac {3} {4}$

It is noted that the shaded area is above the line.

We also observe in the graph that point (0,0) belongs to it.

Finally, it is observed that the intersection with the y-axis is -2. That is: (0, -2)

Answer:

The correct options are: B, D, E

See attached image

Bess [88]3 years ago

5 0

Answer:

its

B........ The graph of y > Three-fourthsx – 2 is a dashed line.

D......One solution to the inequality is (0, 0).

E....The graph intercepts the y-axis at (0, –2).

Step-by-step explanation:

just took the quiz on edge

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erica [24]

Answer:

-3y

Step-by-step explanation:

- $\sqrt{9}$ $\sqrt{y^{2} }$

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5 0

3 years ago

Select the next item in the sequence.

Lapatulllka [165]

9514 1404 393

Answer:

B. 11.000

Step-by-step explanation:

The function looks like a reflected and translated exponential function with a horizontal asymptote near y = 11.000. The rate of change is decreasing so fast that the next value is expected to be very near 10.994. The closest one among the answer choices is 11.000.

_____

First differences are 0.811 and 0.011. The latter is about 0.0136 times the former. At that rate of change, we expect the next first difference to be about 0.000149, which would make the next number in sequence be about 10.9941—very little change from 10.994.

Clearly, first differences are not constant, so the function is not linear. Ratios of the numbers are not constant, so this is not an exponential (geometric) sequence. A reflected exponential function of the type described is a good fit.

With only 3 points given, the rule is not at all obvious. The next term could legitimately be anything you like, and a rule could be made that would fit it.

8 0

3 years ago

Find the surface area and volume of the following figures.

Oduvanchick [21]

White figure Area = 128 $\pi$

White surface area = appro. 301

Yellow area = 320 $\pi$

Yellow Surface area = approx. 653

Area found with = 2πrh+2πr2

Surface Area found with = 2πrh+2πr2

You need to memorize them for tests

Hope that helped!!! k

4 0

4 years ago

Read 2 more answers

Find the volume of the solid generated by revolving the region bounded above by yequals16 cosine x and below by yequals3 secant

Reika [66]

Answer:

$V=\dfrac{256\pi^2}{3}+46\pi\sqrt{3}$

Step-by-step explanation:

Let the functions be $y_1$ and $y_2$

$y_1=3\sec x$

$y_2=16\cos x$

The volume of the solid of revolution generated is given by

$V = \int_\dfrac{-\pi}{3}^\dfrac{\pi}{3} \pi(y_2^2 - y_1^2) dx$

$V = \int_\dfrac{-\pi}{3}^\dfrac{\pi}{3} \pi(16\cos x)^2 - (3\sec x)^2) dx$

$V = \pi\int_\dfrac{-\pi}{3}^\dfrac{\pi}{3} 256\cos^2 x - 9\sec^2x) dx$

Using the trigonometric expansion of $\cos^2x = \dfrac{1+\cos 2x}{2}$ and integrating it as well as using the fact that the integral of $\sec^2x=\tan x$ , we have

$V =\dfrac{256\pi^2}{3}+46\pi\sqrt{3}$

7 0

3 years ago

Hoy many pairs of parallel lines does a Square have

Y_Kistochka [10]

Two pairs because a square has 4 sides

6 0

4 years ago