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UNO [17]
3 years ago
11

A student claims that the function f(x)=x4+kx2+1 is an even function. Which statement best describes the student's claim?

Mathematics
1 answer:
Natali [406]3 years ago
7 0

Answer:

A

Step-by-step explanation:

Suppose x is odd. Then x^4 is also odd. So x^4 + 1 is even. Then for the whole function to be even, kx^2 should also be even. x^2 is obviously odd as x is odd. So to make kx^2 even, k should be even.

Now suppose x is <em>even</em>. Then x^4 + 1 is odd. So to make the function even, kx^2 should also be odd. But x is even, so kx^2 automatically becomes even, and then the whole function becomes odd. So the student's statement doesn't hold when x is even, no matter what the value of k is.

So the statement is true when x is odd and k is even.

Hope it helps and if it does, plzzzz mark me brainliest...

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7x-3y=-12. Name 2 possible answers for (x,y)
anzhelika [568]

Answer:

x = (-12/7 , 0)

y= (0,4)

Step-by-step explanation:

Plug y=0 into the equation and solve the resulting equation 7x=−12 for x

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/7      /7

x = -12/7     and y =0

Plug x=0 into the equation and solve the resulting equation −3y=−12 for y

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6 0
2 years ago
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4 0
2 years ago
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Marina86 [1]

Answer:

Yes

Step-by-step explanation:

The amount of money Albert receives is described by the expression

y = \begin{cases}0.30x  & \quad \text{if } 0 \leq x \leq 49 \\0.40x & \quad \text{if } x > 50\\\end{cases}

The graph is shown below.

To determine if the relation is a function,  we can use the vertical line test:  

If a vertical line crosses the graph more than once in any location, the relation is not a function.

We see that at no place will a vertical line intersect the graph more than once.

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3 0
3 years ago
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Then to choose 2 letters among 6, with no repeat, there are 6x5 = 30 ways

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8 0
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