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makkiz [27]
3 years ago
9

What’s the solution

Mathematics
2 answers:
Nookie1986 [14]3 years ago
6 0

Answer:

x>=12

Step-by-step explanation:

-3/4x + 2<=-7

-3/4x <= -7 -2

-3/4x<=-9

cross multiply

-3x<=-36

dividing throughout by -3

x>=12

nordsb [41]3 years ago
5 0

Answer:

x ≥ 12

Step-by-step explanation:

-3/4x +2 ≤ -7

Subtract 2 from each side

-3/4x +2-2 ≤ -7-2

-3/4x  ≤ -9

Multiply each side by -4/3, remembering to flip the inequality

-3/4x * -4/3 ≥ - 9 *(-4/3)

x ≥ 12

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3 years ago
Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 2, and the given zeros. -
Musya8 [376]

Answer:

A least degree polynomial, having rational coefficients and a leading coefficient of 2, with,-4, 0, 2, and 4 as the zeros of the polynomial is;

f(x) = 2·x⁴ - 4·x³ - 32·x² + 64·x

Step-by-step explanation:

The given parameters of the polynomial are;

The leading coefficient of the polynomial = 2

The zeros of the polynomial = -4, 0, 2, 4

We note that zeros of -4, and 4 gives a factor of the form, (x² - 4²)

For a zero of the polynomial equal to 0, one of the factors of the polynomial is equal to 'x'

To have a leading coefficient of 2, we can add '2' as a factor of the polynomial

Therefore, we can have the factors of the polynomial as follows;

(x² - 4²)·2·x×(x - 2) = 0

From the above equation, using a graphing calculator, we get the following possible polynomial;

(x² - 4²)·2·x×(x - 2) = 2·x⁴ - 4·x³ - 32·x² + 64·x = 0

Therefore, a polynomial, function of least degree that has rational coefficients, a leading coefficient of 2,and the zeros, -4, 0, 2, and 4 is  f(x) = 2·x⁴ - 4·x³ - 32·x² + 64·x.

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Which equation can be used to solve
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Answer:

Value of <em>x1</em><em> </em><em>=</em><em> </em><em>10</em><em> </em><em> </em><em>and</em><em> </em><em>x2</em><em> </em><em>=</em><em> </em><em> </em><em>-</em><em>3</em>

Step-by-step explanation:

when \: we \: multiply \: the \: above \: matrices \\ we \: get \\ 2x1 + 6x2 = 2 \:  \:  \: \:  and \\ 0x1 + 1x2 =  - 3 \\ then \:  \: x2 =  - 3 \\ on \: substituting \: x2 =  - 3 \:\\  in \: first \: equation \:  \\  2x1 + 6( - 3) = 2 \\ 2x1 - 18 = 2 \\ 2x1 = 20 \\ x1 =  \frac{20}{2}  \\ x1 = 10

<em>HAVE A NICE DAY</em><em>!</em>

<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>.</em>

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