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2 years ago

Find the discriminant and the number of real roots for this equation.

Mathematics

1 answer:

GuDViN [60]2 years ago

8 0

Answer:

Step-by-step explanation:

To find the discriminant you do b^2-4(a)(c), which in this case gives you -28. If your discriminant is less than zero, you will have no real roots.

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Who knows this work?

Diano4ka-milaya [45]

Number 13 would be 4/12 because four ounces are vanilla

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2 years ago

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Using the order of operations, what should be done first to evaluate 12 divided by (negative 6) (3) + (negative 2)? Divide 12 by

koban [17]

12 / (-6)(3) + (-2)

First divide 12 by -6

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3 years ago

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Help help help

shtirl [24]

Answer:

I think 60 - 1200 but i am not conform

7 0

3 years ago

I can’t figure out how to use the zeros in the polynomial. Please explain

Temka [501]

Answer:

a) $P (x) = (x + 3) (x-1) (x-4)$

b) $P (x) = (2x + 5) (5x - 4) (x-6)$

c) $P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2$

Step-by-step explanation:

For the question a * you need to find a polynomial of degree 3 with zeros in -3, 1 and 4.

This means that the polynomial P(x) must be zero when x = -3, x = 1 and x = 4.

Then write the polynomial in factored form.

$P (x) = (x + 3) (x-1) (x-4)$

Note that this polynomial has degree 3 and is zero at x = -3, x = 1 and x = 4.

For question b, do the same procedure.

Degree: 3

Zeros: -5/2, 4/5, 6.

The factors are

$x = -\frac{5}{2}\\\\x +\frac{5}{2} = 0\\\\(2x +5) = 0$

---------------------------------------

$x =\frac{4}{5}\\\\x-\frac{4}{5} = 0\\\\(5x-4) = 0$

--------------------------------------

$x = 6\\\\(x-6) = 0$

--------------------------------------

$P (x) = (2x + 5) (5x - 4) (x-6)$

Finally for the question c we have

Degree: 5

Zeros: -3, 1, 4, -1

Multiplicity 2 in -1

$x = -3\\\\(x-3) = 0$

--------------------------------------

$x = 1\\\\(x-1) = 0$

--------------------------------------

$x = 4\\\\(x-4) = 0$

----------------------------------------

$x = -1\\\\(x + 1) = 0$

-----------------------------------------

$P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2$

8 0

2 years ago

A baseball team's total wins and losses for one season is 162. The number of wins is 6 less than 2 times the number of losses. W

aivan3 [116]

Answer:

x + y = 162 and x = 2y - 6

Step-by-step explanation:

Let the total wins be x

Let the total loss be y

If a baseball team's total wins and losses for one season is 162, this can be expressed as;

x + y = 162 ....... 1

If the number of wins is 6 less than 2 times the number of losses, this is expressed as;

2 times the number of losses = 2y

6 less than 2 times the number of losses = 2y - 6

Now, If the number of wins is 6 less than 2 times the number of losses this is expressed as;

x = 2y - 6 .... 2

Hence the required system of equation are;

x + y = 162 and x = 2y - 6

8 0

2 years ago