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worty [1.4K]
2 years ago
13

When mixed together in varying degrees, the three primary additive colors, consisting of ________, can produce all the other col

ors in the spectrum.
Mathematics
1 answer:
timofeeve [1]2 years ago
4 0

Answer:

Red, Blue, Yellow

Step-by-step explanation:

The three primary colors are red, blue, and yellow.

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What is the area of a triangle with vertices at (−4, 1) , ​ (−7, 5) ​ , and ​ (0, 1) ​ ?
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Plot the points on a graph. Connect the dots into a triangle. See that the height of the triangle is from y=5 down to y=1. So the height is 4 units.
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A = (1/2)(4)(4)
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** Distance formula fyi
d² = (X-x)² + (Y-y)²
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3 years ago
15x=3x+120 <br><br> Show answer
inn [45]
15x = 3x + 120 (subtract both sides by 3x)
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5 0
3 years ago
Read 2 more answers
A drawer contains 4 blue socks, 8 black socks, and 10 white
Sindrei [870]

Answer:

7: 2/11

8: 4/11

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5 0
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Can someone please explain this to me? Thanks!
Makovka662 [10]

Answer:  Choice D

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

==========================================================

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

5 0
2 years ago
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