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2 years ago

Which function is quadratic?

Mathematics

1 answer:

AveGali [126]2 years ago

3 0

Answer:

g(x) is a quadratic function ⇒ 2

Step-by-step explanation:

The quadratic function is the function that has 2 as the greatest power of the variable

The form of the quadratic function is f(x) = ax² + bx + c, where a, b, and c are constant

Let us use the information above to solve the question

∵ f(x) = $1.5^{x}$

∵ x is the exponent of the base 1.5

→ That means f(x) is not in the form of the quadratic function

∴ f(x) is not in the form of the quadratic function above

∴ f(x) does not represent a quadratic function

∴ f(x) is not a quadratic function

∵ g(x) = 500x² + 345x

∴ The greatest power of x is 2

→ That means g(x) is in the form of the quadratic function above

∵ g(x) is in the form of the quadratic function above, where a = 500,

b = 345, and c = 0 (constant values)

∴ g(x) represents a quadratic function

∴ g(x) is a quadratic function

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Natasha_Volkova [10]

Answer:

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Step-by-step explanation:

From the information given:

BY applying sphere coordinates:

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The intersection that takes place in the sphere and the cone is:

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Similarly; in the X-Y plane;

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cosФ = 0

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So here; $\dfrac{\pi}{4} \leq \phi \le \dfrac{\pi}{2}$

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$V = \bigg [-cos \phi \bigg]^{\pi/2}_{\pi/4} \bigg [\theta \bigg]^{2 \pi}_{0} \bigg [\dfrac{\rho^3}{3} \bigg ]^{9}_{0}$

$V = [ -0+ \dfrac{1}{\sqrt{2}}][2 \pi -0] [\dfrac{9^3}{3}- 0 ]$

V = 243 $\sqrt{2} \ \pi$

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