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lys-0071 [83]
3 years ago
15

All three sides and all three angles are equal in an equilateral triangle. Could this figure

Mathematics
2 answers:
exis [7]3 years ago
4 0

Answer:

never

Step-by-step explanation:

Any triangle can never be a parallelogram because it only has three sides.

quadrilateral with parallel sides is a parallelogram.

kherson [118]3 years ago
3 0
Never, a parallelogram has to have two pairs of parallel sides. (Four sides) A triangle only has three sides.
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PIT_PIT [208]

Hi!

The Desmos graph is below.

According to the graph, we can see that one of the points is (1, -5).

<u><em>Additional information:</em></u>

In inequalities, each equation will be <u>shaded</u> a certain direction/side, and anything in that shaded region is a solution to the equation. With 2 inequalities, the solutions to the equations will be in the <u><em>double-shaded</em></u> region.

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<em />

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brainly.com/question/16339562

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4 0
2 years ago
What is the slope of the following table?
Bess [88]

Answer:

-  \frac{1}{2}

c)

Step-by-step explanation:

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5 0
3 years ago
Y=x.arctan(x)^1/2. find dy/dx. pls show steps​
Whitepunk [10]

y=x(\arctan x)^{1/2}

Use the product rule first:

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dx}{\mathrm dx}(\arctan x)^{1/2}+x\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}

\dfrac{\mathrm dy}{\mathrm dx}=(\arctan x)^{1/2}+x\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}

Use the chain rule to compute the derivative of (\arctan x)^{1/2}. Let z=(\arctan x)^{1/2} and take u=\arctan x, so that by the chain rule

\dfrac{\mathrm dz}{\mathrm dx}=\dfrac{\mathrm dz}{\mathrm du}\dfrac{\mathrm du}{\mathrm dx}

\dfrac{\mathrm dz}{\mathrm du}=\dfrac{\mathrm du^{1/2}}{\mathrm du}=\dfrac12u^{-1/2}

\dfrac{\mathrm du}{\mathrm dx}=\dfrac{\mathrm d\arctan x}{\mathrm dx}=\dfrac1{1+x^2}

\implies\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}=\dfrac1{2(\arctan x)^{1/2}(1+x^2)}

So we have

\dfrac{\mathrm dy}{\mathrm dx}=(\arctan x)^{1/2}+\dfrac x{2(\arctan x)^{1/2}(1+x^2)}

You can rewrite this a bit by factoring (\arctan x)^{-1/2}, just to make it look neater:

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1{2(\arctan x)^{1/2}}\left(2\arctan x+\dfrac x{1+x^2}\right)

3 0
3 years ago
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