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suter [353]
2 years ago
15

Lelia says that 75% of a number will always be greater than 50% of a number. Complete the inequality to support Lelia's claim an

d one to show that she is incorrect.
Mathematics
1 answer:
worty [1.4K]2 years ago
3 0

Answer:

Let 'x' and 'y' be two different numbers.

Leila says that 75% of a number will always be greater than 50% of a number. The inequality that represents this statement is the following:

0.75x > 0.5y

Let x = 100 and y=200. We have that:

0.75(100) > 0.5(200)

75 > 100 ❌ INCORRECT ❌

Given that we found a case in which 75% of a number is not greater than 50% of a number, we can conclude that Leila's claim is incorrect.

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Answer:

- 2x^5+ 0x^4 +  \frac{x^3}{2} +0x^2+\frac{x}{4}  + 1

Step-by-step explanation:

Given

\frac{x}{4} - 2x^5 + \frac{x^3}{2} + 1

Required

The standard form of the polynomial

The general form of a polynomial is

ax^n + bx^{n-1} + cx^{n-2} +........+ k

Where k is a constant and the terms are arranged from biggest to smallest exponents

We start by rearranging the given polynomial

- 2x^5+ \frac{x^3}{2} +\frac{x}{4}  + 1

Given that the highest exponent of x is 5;

Let n = 5

Then we fix in the missing terms in terms of n

- 2x^5+ 0x^{n-1} +  \frac{x^3}{2} +0x^{n-3}+\frac{x}{4}  + 1

Substitute 5 for n

- 2x^5+ 0x^{5-1} +  \frac{x^3}{2} +0x^{5-3}+\frac{x}{4}  + 1

- 2x^5+ 0x^{4} +  \frac{x^3}{2} +0x^{2}+\frac{x}{4}  + 1

Hence, the standard form of the given polynomial is - 2x^5+ 0x^4 +  \frac{x^3}{2} +0x^2+\frac{x}{4}  + 1

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2 years ago
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2 years ago
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Answer:
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x coordinate of M' = k * x coordinate of M 
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y coordinate of M' = 0.8 * 4 = 3.2
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For point N':
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8 0
2 years ago
How do you solve -9( -4r + 6s ) + 3s - 7(6s - 2r)
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Final result :

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7 0
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