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3 years ago

0,4 Find the roots of the following quadratic equations

Mathematics

1 answer:

Crazy boy [7]3 years ago

7 0

Answer:

b = b2-4ac

= (-7ab)^2 -4 (6a^2)(-3b^2)

= 49a^2b^2 + 72a^2b^2

= 121a^2b^2

Now, x = (7ab +- root 121a^2b^2)÷2(6a²)

= (7ab +- 11ab ) ÷ 12a²

= ( 20ab)/12a² or 4ab/12a²

= 5/3ab or - 1/3ab .

So, these are its roots .

Step-by-step explanation:

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Simplify the expression below.

Ostrovityanka [42]

Answer: OPTION C.

Step-by-step explanation:

Given the following expression:

$4x(3x- 7) - 19x^2$

You can simplify this expression applying these steps:

First step: You must apply Distributive property:

$=12x^2-28x-19x^2$

Second step: You need to add the like terms.

Then, you get:

$=-7x^2-28x$

You can observe that this simplified expression matches with the expression provided in Option C.

3 0

4 years ago

Read 2 more answers

10 ounces is what fraction of a pound?

kicyunya [14]

There are 16 oz. in a pound.
10/16
5/8

7 0

3 years ago

The resultant of two forces acting on a body has a magnitude of 80 pounds. The angles between the resultant and the forces are 2

kifflom [539]

<span>The basic idea is that you form a parallelogram with those two vectors as the two different side lengths another way to see it: start at the tip of one vector and move in the same direction as the other vector (and the same length as the other vector)

</span><span>With any parallelogram, the adjacent angles are supplementary</span>

180-52-20= 108 degrees

5 0

3 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Anna [14]

Answer:

a) 0.4452

b) 0.0548

c) 0.0501

d) 0.9145

e) 6.08 minutes or greater

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.7 minutes

Standard Deviation, σ = 0.50 minutes.

We are given that the distribution of length of the calls is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(calls last between 4.7 and 5.5 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{4.7 - 4.7}{0.50} \leq z \leq \displaystyle\frac{5.5-4.7}{0.50}) = P(0 \leq z \leq 1.6)\\\\= P(z \leq 1.6) - P(z$

$P(4.7 \leq x \leq 5.5) = 44.52\%$

b) P(calls last more than 5.5 minutes)

$P(x > 5.5) = P(z > \displaystyle\frac{5.5-4.7}{0.50}) = P(z > 1.6)\\\\P( z > 1.6) = 1 - P(z \leq 1.6)$

Calculating the value from the standard normal table we have,

$1 - 0.9452 = 0.0548 = 5.48\%\\P( x > 5.5) = 5.48\%$

c) P( calls last between 5.5 and 6 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{5.5 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(1.6 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$

$P(5.5 \leq x \leq 6) = 5.01\%$

d) P( calls last between 4 and 6 minutes)

$P(4 \leq x \leq 6) = P(\displaystyle\frac{4 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(-1.4 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$