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2 years ago

Consider the equation below.

Mathematics

1 answer:

Marta_Voda [28]2 years ago

7 0

Be because I did this it’s just extremely hard to explain how to trust OK open helps

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Which image shows a plane slicing a double-napped cone to produce a hyperbola?<br><br> Answer is D

Butoxors [25]

The <u>second image</u> in the diagram is a hyperbola. As can be seen, the plane cutting the cone can be at any angle but never equal to the slant angle of the cone. This has a very important implication. The plane cuts both cones of the double-napped cone. The third double-napped cone of Figure 3 shows two hyperbolas.

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1 year ago

IM GOING TO GIVE 251 POINTS TO WHOEVER ANSWER THIS PROBLEM.

Mariulka [41]

Answer:

(x^87*(x^2+4))/54

Step-by-step explanation:

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1 year ago

In ΔABC, if m∠A = 43° and m∠C = 66°, then find m∠B.

weqwewe [10]

The measures of the angles in a triangle add up to 180°.
$m \angle A + m \angle B + m \angle C = 180^\circ \\
m \angle A=43^\circ, \ m \angle C = 66^\circ \\ \downarrow \\
43^\circ+ m \angle B + 66^\circ = 180^\circ \\
m \angle B + 109^\circ = 180^\circ \\
m \angle B =180^\circ - 109^\circ \\
\boxed{m \angle B=71^\circ}$

6 0

2 years ago

Can someone please explain this to me? Thanks!

Makovka662 [10]

Answer: Choice D

$\displaystyle F\ '(x) = 2x\sqrt{1+x^6}\\\\$

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Explanation:

Let g(t) be the antiderivative of $g'(t) = \sqrt{1+t^3}$ . We don't need to find out what g(t) is exactly.

Recall by the fundamental theorem of calculus, we can say the following:

$\displaystyle \int_{a}^{b} g'(t)dt = g(b)-g(a)$

This theorem ties together the concepts of integrals and derivatives to show that they are basically inverse operations (more or less).

So,

$\displaystyle F(x) = \int_{\pi}^{x^2}\sqrt{1+t^3}dt\\\\ \displaystyle F(x) = \int_{\pi}^{x^2}g'(t)dt\\\\ \displaystyle F(x) = g(x^2) - g(\pi)\\\\$

From here, we apply the derivative with respect to x to both sides. Note that the $g(\pi)$ portion is a constant, so $g'(\pi) = 0$

$\displaystyle F(x) = g(x^2) - g(\pi)\\\\ \displaystyle F \ '(x) = \frac{d}{dx}[g(x^2)-g(\pi)]\\\\\displaystyle F\ '(x) = \frac{d}{dx}[g(x^2)] - \frac{d}{dx}[g(\pi)]\\\\ \displaystyle F\ '(x) = \frac{d}{dx}[x^2]*g'(x^2) - g'(\pi) \ \text{ .... chain rule}\\\\$

$\displaystyle F\ '(x) = 2x*g'(x^2) - 0\\\\ \displaystyle F\ '(x) = 2x*g'(x^2)\\\\ \displaystyle F\ '(x) = 2x\sqrt{1+(x^2)^3}\\\\ \displaystyle F\ '(x) = \boldsymbol{2x\sqrt{1+x^6}}\\\\$

Answer is choice D

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1 year ago

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sukhopar [10]

Answer:

Step-by-step explanation:

because I am right

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2 years ago

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