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3 years ago

Find two numbers that have 2, 3, and 5 as factors. What other factors do those two numbers have in common?

1 answer:

4 0

To find the first number, just multiply the three numbers together.
$2*3*5 \\ 6*5 \\ 30$

To find the second number, multiply the three numbers together along with another number (this number can be anything).
$2*3*5*4 \\ 6*5*4 \\ 30*4 \\ 120$

What They Have in Common:
- All numbers will be even, since you are multiplying by 2.
- All numbers will end in a zero, since you are multiplying by 5 and also 2.

You might be interested in

Brass is made up of copper and zinc. 100 grams of brass contains 20 grams of zinc.

Nana76 [90]

Answer:

A. 12 grams Zn

B. 280 g of copper.

Step-by-step explanation:

100 grams of brass

20 grams of zinc

100 grams of brass has 20 grams of Zn.

20/100 = 0.200 Zn 20% Zn and 80% copper.

A. (60 g brass)*(0.2) = 12 grams Zn

B. (350 g brass)*(0.80) = 280 g of copper.

5 0

2 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

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

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

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

$R_3\rightarrow R_3-2R_1$

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

$R_3\rightarrow R_3-R_2$

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

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago

Rationalise the denominator by 6 root 7 minus 1

nikklg [1K]

Answer:

Assuming the numerator was 1.

$(6\sqrt{7}+1)/251$

Step-by-step explanation:

Assuming the numerator was 1.

$1/(6\sqrt{7} -1)\\ (6\sqrt{7}+1)/((6\sqrt{7} - 1)(6\sqrt{7} + 1))\\ (6\sqrt{7}+1)/((6\sqrt{7})^2 + 6\sqrt{7} - 7\sqrt{7} - 1)\\(6\sqrt{7}+1)/((6\sqrt{7})^2 - 1)\\(6\sqrt{7}+1)/((36*7 - 1)\\(6\sqrt{7}+1)/251$

7 0

3 years ago

Which of the following about the normal distribution is not true?

Lostsunrise [7]

Answer:

Normal distribution is a Continuous distribution.

Step-by-step explanation:

Normal Distribution:

The following information is true about normal distribution.

Theoretically, the mean, median, and mode are the same.

Its parameters are the mean, mu, and standard deviation, sigma.

About two thirds of the observations fall within plus or minus 1 standard deviation from the mean.

But normal distribution is not a discrete distribution. It is a continuous distribution.

The normal distribution is a continuous probability distribution. This has several implications for probability.
The probability that a normal random variable X equals any particular value is 0.

5 0

3 years ago

Solve the given equation by completing the square.

Dmitriy789 [7]

Answer:

a = -4, b = 3 and c = 6 (OR)

a = -4, b = -3 and c = 6

Step-by-step explanation:

$x^{2} +8x = 38$