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ra1l [238]
3 years ago
15

Evaluate 7-5p+3q when p=1 and q=7

Mathematics
2 answers:
skad [1K]3 years ago
4 0

Answer:

Rewrite the equation with the values instead of letters.

7-(5 x 1) + (3 x 7)

7-5+21

7-5 = 2

2 + 21 = 23

23 is your answer.

Step-by-step explanation:

KIM [24]3 years ago
3 0

Answer:

23

Step-by-step explanation:

<em><u>Since it gives you the values of p and q, just plug it into the expression:</u></em>

7 - 5p + 3q

7 - 5(1) + 3(7)

7 - 5 + 21

2 + 21

23

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Find possible zeroes<br> f(x)=3x^6+4x^3-2x^2+4
larisa [96]

The possible zeros of f(x) = 3x^6 + 4x^3 -2x^2 + 4 are \mathbf{\pm\{1,2,4,\frac 13, \frac 23,\frac{4}{3}}\}

<h3>How to determine the possible zeros?</h3>

The function is given as:

f(x) = 3x^6 + 4x^3 -2x^2 + 4

The leading coefficient of the function is:

p = 3

The constant term is

q = 4

Take the factors of the above terms

p = 1 and 3

q = 1, 2 and 4

The possible zeros are then calculated as:

\mathbf{Zeros = \pm\frac{Factors\ of\ q}{Factors\ of\ p}}

So, we have:

\mathbf{Zeros = \pm\frac{1,2,4}{1,3}}

Expand

\mathbf{Zeros = \pm\frac{1,2,4}{1},\pm\frac{1,2,4}{3}}

Solve

\mathbf{Zeros = \pm\{1,2,4,\frac 13, \frac 23,\frac{4}{3}}\}

Hence, the possible zeros of f(x) = 3x^6 + 4x^3 -2x^2 + 4 are \mathbf{\pm\{1,2,4,\frac 13, \frac 23,\frac{4}{3}}\}

Read more about rational root theorem at:

brainly.com/question/9353378

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