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3 years ago

How do you do 3x (3X+6) please list how to do in order please. thanks!

2 answers:

Nadusha1986 [10]3 years ago

8 0

Do what? If you mean to find when it equals zero...

By inspection you can see that if x=0 or 2 the expression is equal to zero...

You multiply each term in the parentheses by the term outside the parentheses...like:

3x*3x+3x*6 which is equal to:

9x^2+18x If you meant to expand it....

9x(x+2) If you meant to factor it completely

Free_Kalibri [48]3 years ago

8 0

3x(3x+6)
=(3x)(3x+6)
=(3x)(3x)+(3x)(6)
=9x^2+18x-- final answer

Cheers,Irys

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The Venn diagram shows the number of students that exercise in different ways. What number should go in the highlighted box belo

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Answer:

Step-by-step explanation:

Add up all the numbers

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3 years ago

I'm having trouble solving this I don't know the steps

Andreas93 [3]

Answer:

x=4

Step-by-step explanation:

21x+6+90=180

First, we need to subtract the numbers.

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21x=84

Now, we need to find the value for 1x, so divide the whole equation by 21

84/21

21/21

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Hope this helps!

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Zepler [3.9K]

Answer:

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3 years ago

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

LenaWriter [7]

Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let X = number of vehicles that pass the inspection.

The probability of the random variable X is P (X) = 0.70.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

$=(0.70)^{3}\\=0.343$

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

$=1-0.343\\=0.657$

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

= P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

+ P (Only 3rd vehicle passes)

$=(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189$

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

+ P (0 vehicles passes)

$=0.189+(0.30\times0.30\times0.30)\\=0.216$

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let X = all 3 vehicle passes and Y = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

$P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525$

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

7 0

3 years ago