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2 years ago

Yuri thinks that 3/4 is a root of the following function. q(x) = 6x3 + 19x2 – 15x – 28 Explain to Yuri why 3/4 cannot be a root.

2 answers:

6 0

If $\frac{3}{4}$ is a root of the equation, q(x) = 6x3 + 19x2 – 15x – 28 then by the factor formula q( $\frac{3}{4}$ ) = 0; if that is not the case then $\frac{3}{4}$ cannot be a factor of the function.

Now, q( $\frac{3}{4}$ ) = 6( $\frac{3}{4}$ )³ + 19( $\frac{3}{4}$ )² – 15( $\frac{3}{4}$ ) – 28
= $- \frac{833}{32}$

Therefore Yuri, since q( $\frac{3}{4}$ ) ≠ 0 then it implies that $\frac{3}{4}$ is not a root of q(x).

Additionally, a root should be expressed in terms of x, thus the possible root ought to be written as x = $\frac{3}{4}$ , instead of just $\frac{3}{4}$

valina [46]2 years ago

6 0

Answer:

According to the rational root theorem, potential rational roots must be in p/q form where p is a factor of the constant term and q is a factor of the leading coefficient.

3 is not a factor of 28, and 4 is not a factor of 6. So, 3/4 not satisfy the rational root theorem.

Substituting 3/4 in for x does not result in 0. The remainder is not 0 when dividing by

x –3/4

Step-by-step explanation:

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Grasshoppers are distributed at random in a large field according to a Poisson process with parameter a 5 2 per square yard. How

HACTEHA [7]

In this question, the Poisson distribution is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Parameter of 5.2 per square yard:

This means that $\mu = 5.2r$ , in which r is the radius.

How large should the radius R of a circular sampling region be taken so that the probability of finding at least one in the region equals 0.99?

We want:

$P(X \geq 1) = 1 - P(X = 0) = 0.99$

Thus:

$P(X = 0) = 1 - 0.99 = 0.01$

We have that:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-5.2r}*(5.2r)^{0}}{(0)!} = e^{-5.2r}$

Then

$e^{-5.2r} = 0.01$

$\ln{e^{-5.2r}} = \ln{0.01}$

$-5.2r = \ln{0.01}$

$r = -\frac{\ln{0.01}}{5.2}$

$r = 0.89$

Thus, the radius should be of at least 0.89.

Another example of a Poisson distribution is found at brainly.com/question/24098004

3 0

2 years ago

What is the first step in solving the equation?<br><br> 14 = -2m

ollegr [7]

Answer:

first you need to divide each side by -2

Step-by-step explanation:

start with dividing by -2, then you are left with -7 = m which is the answer.

7 0

1 year ago

There is a parallelogram ABCD with diagonals AC and BD. The diagonals AC and BD intersects each other at point E. Side AB is con

grandymaker [24]

Answer:

SAS theorem

Step-by-step explanation:

Given

$\square ABCD$

$\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD$

$\angle BAC = \angle DCA$

Required

Which theorem shows △ABE ≅ △CDE.

From the question, we understand that:

AC and BD intersects at E.

This implies that:

$\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC$

and

$\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED$

So, the congruent sides and angles of △ABE and △CDE are:

$\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD$ ---- S

$\angle BAC = \angle DCA$ ---- A

$\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED$ or $\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC$ --- S

<em>Hence, the theorem that compares both triangles is the SAS theorem</em>

4 0

1 year ago

How can I solve this? I need help its for geometry

Scilla [17]

What is the question

4 0

2 years ago

Which represents the solution set of an any quality 5X-9 greater than 21

Svet_ta [14]

Answer:

The last one

Step-by-step explanation:

5x<30

x<6

Because when you divide by pos+ number

the symbol cannot change.

6 0

2 years ago