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3 years ago

3x^3 - 2y^-6x^2y^2+xy for x=2/3 and y=1/2

Mathematics

1 answer:

Dominik [7]3 years ago

4 0

Answer:

-13

Step-by-step explanation:

3x^3 - 2y^-6x^2y^2+xy

x=2/3 and y=1/2

Solve :

3x^3 - 2y^-6x^2y^2+xy

Put value of x and y in this equation

= 3(2/3)^3 - 2(1/2)^-6(2/3)^2(1/2)^2+(2/3)(1/2)

= -13

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Answer:

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Vladimir [108]

Answer:

$cos^2\theta + sin^2\theta = 1$

Step-by-step explanation:

Given

$(\frac{b}{r})^2 + (\frac{a}{r})^2$

Required

Use the expression to prove a trigonometry identity

The given expression is not complete until it is written as:

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = (\frac{r}{r})^2$

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Having said that:

The expression can be further simplified as:

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4 years ago

Could someone help me rnnn?

GuDViN [60]

Answer:

vertex = (0, -4)

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Step-by-step explanation:

Given:

y-intercept of parabola: -4
parabola passes through points: (-2, 8) and (1, -1)

Vertex form of a parabola: $y=a(x-h)^2+k$

(where (h, k) is the vertex and $a$ is some constant)

Substitute point (0, -4) into the equation:

$\begin{aligned}\textsf{At}\:(0,-4) \implies a(0-h)^2+k &=-4\\ah^2+k &=-4\end{aligned}$

Substitute point (-2, 8) and $ah^2+k=-4$ into the equation:

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