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Nitella [24]
3 years ago
9

Very Easy Question 2/3 divided by 5.

Mathematics
1 answer:
Stella [2.4K]3 years ago
6 0

Answer:

2/15

Step-by-step explanation:

2/3 divided by 5

= 2/15

<em>Hope this helps</em>

<em>-Amelia</em>

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Can you help with this? Thank you!
Gnom [1K]

To solve for j, cross multiply:

4 x 45 = 18 x j

180 = 18j

Divide both sides by 18:

j = 10

5 0
3 years ago
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In how many ways can 15 basketball players be listed in a program?
Viktor [21]
The answer is 15 !

Explanation:
You have 15 slots, each of which will hold one of the names of the players. You need to fill the first slot with some name.
7 0
4 years ago
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A television newscaster from a 24-hour news station wants to estimate the number of hours per day its viewers watch
topjm [15]

Answer:

This isn't the best worded question but from what I understand a larger sample size decreases the margin of error. If he would like a more accurate answer a larger sample size of the viewers will give more accurate answers.

Step-by-step explanation:

4 0
3 years ago
Using addition formula solve tan 15​
salantis [7]

Answer:

2 - \sqrt{3}

Step-by-step explanation:

Using the addition formula for tangent

tan(A - B) = \frac{tanA-tanB}{1+tanAtanB} and the exact values

tan45° = 1 , tan60° = \sqrt{3} , then

tan15° = tan(60 - 45)°

tan(60 - 45)°

= \frac{tan60-tan45}{1+tan60tan45}

= \frac{\sqrt{3}-1 }{1+\sqrt{3} }

Rationalise the denominator by multiplying numerator/ denominator by the conjugate of the denominator.

The conjugate of 1 + \sqrt{3} is 1 - \sqrt{3}

= \frac{(\sqrt{3}-1)(1-\sqrt{3})  }{(1+\sqrt{3})(1-\sqrt{3})  } ← expand numerator/denominator using FOIL

= \frac{\sqrt{3}-3-1+\sqrt{3}  }{1-3}

= \frac{-4+2\sqrt{3} }{-2}

= \frac{-4}{-2} + \frac{2\sqrt{3} }{-2}

= 2 - \sqrt{3}

3 0
3 years ago
Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

And finally, we can eliminate the second 1 by adding -(R₃):

\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
3 years ago
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