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3 years ago

Please help me look at the picture

Mathematics

1 answer:

harina [27]3 years ago

8 0

The answer is choice A.

We're told that the left and right walls of the cube (LMN and PQR) are parallel planes. Any line contained in one of those planes will not meet another line contained in another plane. With choice A, it's possible to have the front and back walls be non-parallel and still meet the initial conditions. If this is the case, then OS won't be paralle to NR. Similarly, LP won't be parallel to MQ.

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The measures of two complementary angles are on the ratio 2:3. What is the measure of the smaller angle?

notsponge [240]

the smaller angle is 2x = 2 times 18 = 36 degrees.
the larger angle is 3x = 3x = 3 * 18 = 54 degrees.

8 0

3 years ago

The tubes in a solar collector are filled with _____.

Olegator [25]

The tubes in a solar collector are filled with fluid.

4 0

4 years ago

Read 2 more answers

Please help me on this question if you would want brianleist!! Tank yah!! ^^

harkovskaia [24]

False

Explanation: Coordinate Planes are made of 2 intersecting perpendicular lines.

5 0

3 years ago

Identify the relative maximum value of g(x) for the function shown below g(x) = 2/ x^2 +3

Dahasolnce [82]

Answer:

The relative maximum value is $\frac{2}{3}$

Step-by-step explanation:

The given function is

$g(x)=\frac{2}{x^2+3}$

We differentiate to obtain;

$g'(x)=-\frac{4x}{(x^2+3)^2}$

At turning points $g'(x)=-\frac{4x}{(x^2+3)^2}=0$

$\implies x=0$

$g''(x)=\frac{16x^2}{(x^2+3)^3}- \frac{4}{(x^2+3)^2}$

We apply the second derivative test to obtain:

$g''(0)=\frac{16(0)^2}{((0)^2+3)^3}- \frac{4}{((0)^2+3)^2}=-\frac{4}{9}$

Since the second derivative is negative, there is a relative maximum at x=0.

We substitute x=0 into the original function to obtain the relative maximum value.

$g(0)=\frac{2}{(0)^2+3}=\frac{2}{3}$

7 0

3 years ago

Read 2 more answers

The complex fourth roots of 5-5sqrt3i

melamori03 [73]

First convert the complex number to polar form

|5 - 5sqrt3i| = 25sqrt10

argument = arctan - sqrt3 = -pi/3

so 5 - 5sqrt3i = 25sqrt10(cos(-pi/3) + i sin(-pi/3))

the angle is in the 4th quadrant so we could write it as 2pi-pi/3 = 5p/3

= 25sqrt10(cos 5pi/3 + i sin 5pi/3)

Now if r(cosx + isin x) is a 5th root of 5-5sqrt3i then

r^5(cos x + i sinx)^5 = 25sqrt10(cos 5pi/3 + i sin 5pi/3)

r^5 = 25sqrt10 and cos5x + i sin 5x = cos 5pi/3 + i sin 5pi/3

i have to go urgently so i have to leave it to you to finish this

3 0

3 years ago