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3 years ago

Which statements about the graph of the function f(x) = 2x2 – x – 6 are true? Select two options.

Mathematics

2 answers:

luda_lava [24]3 years ago

7 0

Answer:

D and C

Step-by-step explanation:

E2020

natka813 [3]3 years ago

5 0

Answer:

D and C

Step-by-step explanation:

Took the quiz

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Heres the answer l.m.a.o

torisob [31]

Answer:

That's a solid answer there.

Step-by-step explanation:

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3 years ago

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Which diagram shows the medians of a triangle?

aalyn [17]

For this case we have that, by definition, the medians of a triangle are the line segments that are drawn from a vertex to the midpoint of the opposite side of the vertex. The medians of a triangle are concurrent at a point.

If we look at the figure, we can discard triangles 3 and 4.

Also, by definition, the median divides the vertex angle into two equal angles.

Thus, the correct option is triangle 1.

Answer:

Triangle 1

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3 years ago

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Rectangle ABCD is dilated by a scale factor of 1/2 with a center of dilation at the origin. What are the coordinates of the imag

Alex Ar [27]

A. (-2,3) because Rectangle ABCD is dilated by a scale factor of 1/2.

6 0

3 years ago

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Less common denominators for 7/10 2/5

Lera25 [3.4K]

The LCD is 10 because you can multiply 2/5 by 2 to get 4/10

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3 years ago

How to differentiate ?

Bas_tet [7]

Use the power, product, and chain rules:

$y = x^2 (3x-1)^3$

• product rule

$\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d(x^2)}{\mathrm dx}\times(3x-1)^3 + x^2\times\dfrac{\mathrm d(3x-1)^3}{\mathrm dx}$

• power rule for the first term, and power/chain rules for the second term:

$\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(x-1)^2\times\dfrac{\mathrm d(3x-1)}{\mathrm dx}$

• power rule

$\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(3x-1)^2\times3$

Now simplify.

$\dfrac{\mathrm dy}{\mathrm dx} = 2x(3x-1)^3 + 9x^2(3x-1)^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2 \times (2(3x-1) + 9x) \\\\ \boxed{\dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2(15x-2)}$

You could also use logarithmic differentiation, which involves taking logarithms of both sides and differentiating with the chain rule.

On the right side, the logarithm of a product can be expanded as a sum of logarithms. Then use other properties of logarithms to simplify

$\ln(y) = \ln\left(x^2(3x-1)^3\right) \\\\ \ln(y) = \ln\left(x^2\right) + \ln\left((3x-1)^3\right) \\\\ \ln(y) = 2\ln(x) + 3\ln(3x-1)$

Differentiate both sides and you end up with the same derivative:

$\dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac2x + \dfrac9{3x-1} \\\\ \dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \times x^2(3x-1)^3 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(15x-2)(3x-1)^2$

7 0

2 years ago