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3 years ago

Write a polynomial function of least degree with integral coefficients that has the given zeros. -5, 3i

Mathematics

1 answer:

Virty [35]3 years ago

5 0

Solving for the polynomial function of least degree with integral coefficients whose zeros are -5, 3i

We have:
x = -5

Then x + 5 = 0

Therefore one of the factors of the polynomial function is (x + 5)

Also, we have:
x = 3i
Which can be rewritten as:
x = Sqrt(-9)
Square both sides of the equation:
x^2 = -9
x^2 + 9 = 0

Therefore one of the factors of the polynomial function is (x^2 + 9)

The polynomial function has factors: (x + 5)(x^2 + 9)
= x(x^2 + 9) + 5(x^2 + 9)

= x^3 + 9x + 5x^2 = 45

Therefore, x^3 + 5x^2 + 9x – 45 = 0

f(x) = x^3 + 5x^2 + 9x – 45

The polynomial function of least degree with integral coefficients that has the given zeros, -5, 3i is f(x) = x^3 + 5x^2 + 9x – 45

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Which of the following numbers is an integer 1.33333..., 5.2, 2/3,-3

kap26 [50]

Integers are whole numbers so the answer is -3

7 0

3 years ago

An arched drainage culvert can be modelled by the function:

Zanzabum

Substitute 136.5 for y.

$136.5=\frac{x(191-x)}{60}$

$8190=x(191-x)$

$8190=191x-x^{2}$

$x^{2} -191x+8190=0$

Compare this equation with $ax^{2} +bx+c=0$

a = 1, b = - 191, c = 8190

$x=\frac{-b+\sqrt{b^{2}-4ac } }{2}$ or

$x=\frac{-b-\sqrt{b^{2}-4ac } }{2}$

$x=\frac{191+\sqrt{191^{2}-4(1)(8190) } }{2}$ or

$x=\frac{191-\sqrt{191^{2}-4(1)(8190) } }{2}$

$x=\frac{191+\sqrt{36481-32760 } }{2}$ or

$x=\frac{191-\sqrt{36481-32760 } }{2}$

$x=\frac{191+\sqrt{3721 } }{2}$ or

$x=\frac{191-\sqrt{3721 } }{2}$

$x=\frac{191+61}{2}$ or

$x=\frac{191-61}{2}$

$x=\frac{252}{2}$ or

$x=\frac{130}{2}$

x = 126 or x = 65

Hence, the points on the curve are (65, 136.5) and (126, 136.5).

The width of the air space is the distance between these points.

Width = $\sqrt{(x_{2}- x_{1})^{2} +( y_{2}- y_{1}) ^{2}$

= $\sqrt{(126- 65)^{2} +(136.5-136.5) ^{2}$

= 126 - 65

= 61

Hence, width of the air space is 61m.

5 0

2 years ago

Evaluate the expression if m = 3.<br><br> 3m^2<br><br><br> (SHOW YOUR WORK)

iragen [17]

The value of given expression when m = 3 is 27

<h3><u>Solution:</u></h3>

Given expression is $3m^2$

We have to evaluate the given expression for m = 3

To find for m is equal to 3, substitute m = 3 in given expression

From given expression,

$\rightarrow 3m^2$

Plug in m = 3 in above expression

$\rightarrow 3(3)^2$ ------ eqn 1

We know that,

$a^2$ can be expanded as,

$a^2=a \times a$

Applying this in eqn 1, we get

$\rightarrow 3(3)^2=3 \times (3 \times 3)$

Simplify the above expression

$\rightarrow 3(3)^2=3 \times (3 \times 3) = 3 \times 9 = 27$

Therefore, for m = 3 we get,

$\rightarrow 3(3)^2=27$

Thus value of given expression when m = 3 is found

5 0

3 years ago

Please help me answer these in your own words!!!

liubo4ka [24]

6. any thang less than 90 degress

7. only 90 degress

8.any thang above 90 but not 180

9. only 180

10. the sum of 2 angels has to make 90 degress

11. the sum of 2 angles has to make 180 degress

4 0

2 years ago

The bus travels 3 hours going 35 miles per hour. How far does the bus travel in that amount of time? *

3241004551 [841]

The answer is: " 105 mi. " .
____________________________________________

Explanation:
____________________________________________
Note that: "that amount of time" is "3 hours".

→ $\frac{35 mi}{h} * \frac{3h}{1}$ = ? ;

→ The units of "h" (for "hour/s") cancel out to "1" ; since ("h/h = 1") ;

→ and we are left with (35 * 3) mi. = 105 mi.
____________________________________________
The answer is: " 105 mi. " .
____________________________________________

3 0

3 years ago