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2 years ago

Which cannot be a cross section of the figure? A cross section of the figure cannot be a ____

Mathematics

2 answers:

vichka [17]2 years ago

5 0

Answer:

it is circle

Step-by-step explanation:

just got it correct

Lunna [17]2 years ago

4 0

Answer:

the answer is circle

Step-by-step explanation:

i got it right on edge2020 :)

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Using the following image, find the. midpoint of the line by completing the problems attached.

enot [183]

Answer:4/1 (I think)

Step-by-step explanation:

It’s rise over run, so if you look at it what’s the rise what’s the run, of the middle point

5 0

2 years ago

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Which expression is equivalent to -4(7-2m)+11m

dimulka [17.4K]

Answer:b

Step-by-step explanation:

8 0

2 years ago

You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta

vladimir1956 [14]

Answer:

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

Step-by-step explanation:

The slope of the tangent line at a point of the curve is:

$m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$m = -\frac{\cos 2\theta}{\sin \theta}$

The tangent line is horizontal when $m = 0$ . Then:

$\cos 2\theta = 0$

$2\theta = \cos^{-1}0$

$\theta = \frac{1}{2}\cdot \cos^{-1} 0$

$\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right)$ , for all $i \in \mathbb{N}_{O}$

$\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}$ , for all $i \in \mathbb{N}_{O}$

The first four solutions are:

x: $\sqrt{2}$ $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$

y: 1 -1 1 -1

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

6 0

2 years ago

SOMEONE PLEASE HELP.>>>>>>

alexira [117]

Answer:

The answer would be B

6 0

3 years ago

Find the inverse of the function y = 2x2 + 2

Anvisha [2.4K]

We have that
y = 2x²<span> + 2

step 1

</span><span>exchange the value of x for y and the value of y for x
</span>y = 2x² + 2------> x=2y²+2

step 2
clear y variable
x=2y²+2----> 2y²=x-2----> y²=[(x-2)/2]-----> y=(+/-)√[(x-2)/2]

step 3
the inverse is
f(x)-1=(+/-)√[(x-2)/2]

8 0

2 years ago

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