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2 years ago

If 2x+5=8x then 12x=?

Mathematics

2 answers:

vichka [17]2 years ago

8 0

Answer:

The value of 12x is <u>10</u>

Step-by-step explanation:

We need to find the value of 12x

The given expression is 2x + 5 = 8x

Subtract 2x from both the sides in above expression,

2x + 5 - 2x = 8x - 2x

5 = 6x

Multiply both the sides by '2' in above expression

2(5) = 2(6x)

10 = 12x

Hence, the value of 12x is <u>10</u>

UkoKoshka [18]2 years ago

5 0

2x+5=8x

5=6x

x=5/6

12(5/6)

x=10

Guest

1 year ago

Why did you multiply by 2? Is it because you were at 6x and needed to be at 12x? If so, does that mean if you were at 6x and needed to be at 24x, you would’ve multiplied by 4 instead of 2?

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2 years ago

Read 2 more answers

Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang

kkurt [141]

Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

$P=1\cdot\dfrac{9}{11}\cdot\dfrac{6}{10}\cdot\dfrac{3}{9}=\dfrac{162}{990}\approx0.164$

c) We can repeat the method for the previous probabilty.

The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

$P=2\cdot\left(1\cdot\dfrac{2}{11}\right)\cdot\left(\dfrac{9}{10}\cdot\dfrac{2}{9}\right)+\left(1\cdot\dfrac{9}{11}\cdot\dfrac{4}{10}\cdot\dfrac{2}{9}\right)\\\\\\P=2\cdot\dfrac{36}{990}+\dfrac{72}{990}=\dfrac{144}{990}\approx0.145$

5 0

2 years ago

Find all x in set of real numbers R Superscript 4 that are mapped into the zero vector by the transformation Bold x maps to Uppe

sukhopar [10]

Answer:

$x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]$

Step-by-step explanation:

According to the given situation, The computation of all x in a set of a real number is shown below:

First we have to determine the $\bar x$ so that $A \bar x = 0$

$\left[\begin{array}{cccc}1&-3&5&-5\\0&1&-3&5\\2&-4&4&-4\end{array}\right]$

Now the augmented matrix is

$\left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\2&-4&4&-4\ |\ 0\end{array}\right]$

After this, we decrease this to reduce the formation of the row echelon

$R_3 = R_3 -2R_1 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&2&-6&6\ |\ 0\end{array}\right]$

$R_3 = R_3 -2R_2 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]$

$R_2 = 4R_2 +5R_3 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&4&-12&0\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]$

$R_2 = \frac{R_2}{4}, R_3 = \frac{R_3}{-4} \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&1\ |\ 0\end{array}\right]$

$R_1 = R_1 +3 R_2 \rightarrow \left[\begin{array}{cccc}1&0&-4&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]$

$R_1 = R_1 +5 R_3 \rightarrow \left[\begin{array}{cccc}1&0&-4&0\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]$

$= x_1 - 4x_3 = 0\\\\x_1 = 4x_3\\\\x_2 - 3x_3 = 0\\\\ x_2 = 3x_3\\\\x_4 = 0$

$x = \left[\begin{array}{c}4x_3&3x_3&x_3\\0\end{array}\right] \\\\ x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]$

By applying the above matrix, we can easily reach an answer

5 0

2 years ago

Is it possible to form a square whose area is 18 by connecting four lattice points? Explain.

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2 years ago

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anzhelika [568]

Answer:

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Step-by-step explanation:

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