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3 years ago

What is the product of? (7x^2)(2x^3+5)(x^2-4x-9)

2 answers:

6 0

$(7x^{2})(2x^{3} +5)( x^{2} -4x-9)
$
$(14 x^{5} +35 x^{2})( x^{2} -4x-9)$
= $14 x^{7} -56 x^{6}-126 x^{5}+35 x^{4}-140 x^{3}-315 x^{2}$

devlian [24]3 years ago

5 0

Answer:

$14x^7 -56x^6-126x^5+35x^4-140x^3-315x^2$

Step-by-step explanation:

(7x^2)(2x^3+5)(x^2-4x-9)

Find the product of $(7x^2)(2x^3+5)(x^2-4x-9)$

Multiply the parenthesis

$(7x^2)(2x^3+5)=14x^5+35x^2$

Now we multiply 14x^5+35x^2 inside last parenthesis

$( 14x^5+35x^2)(x^2-4x-9)$

$14x^7 -56x^6-126x^5+35x^4-140x^3-315x^2$

Combine like terms if any

$14x^7 -56x^6-126x^5+35x^4-140x^3-315x^2$

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Ms. Reynold's sprinkler system has 5 stations that water all parts of her front and back lawn. Each station runs for an equal am

mina [271]

Answer: 40 minutes

Step-by-step explanation:

It takes 16 minutes for the first 2 stations to water their sections which means that the time taken per station is:

= 16/2

= 8 minutes per station

Ms. Reynolds has 5 stations.

Total time taken will therefore be:

= 8 * 5

= 40 minutes

3 0

3 years ago

Please help me with this question, I'm stuck ;( .

sveticcg [70]

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7 0

3 years ago

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

3 years ago

The ratio to isabella's money to shane's money is 3:11 if isabella has $33 how much money do shane and isabella have together us

alexgriva [62]

Shane and Isabella have 154 dollars

4 0

3 years ago

Standard deviation of 8, 4, 14, 16, 8

Ivan

Answer:

The answer for standard deviation would be:

4.382

3 0

3 years ago