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marissa [1.9K]
3 years ago
14

Which function is quadratic?

Mathematics
1 answer:
AveGali [126]3 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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Answer:

Step-by-step explanation:

Property Rhombus

1 All sides are congruent ✓

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3 0
3 years ago
Find x such that -1/3 = 7/x
Setler [38]

Answer:

x= -21

Step-by-step explanation:

-1/3=7/x

-1x=21

-1x/-1=21/-1

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3 years ago
How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?
BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:  

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}

Where n = 8 and r = 0,1....8

When r = 0

\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1

When r = 1

\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8

When r = 2

\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8

When r = 3

\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56

When r = 4

\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70

When r = 5

\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56

When r = 6

\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28

When r = 7

\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8

When r = 8

\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

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Ostrovityanka [42]

Answer:

D

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It is a trinomial with a degree of 3.

This is the correct answer on the exam.

7 0
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