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Brums [2.3K]
3 years ago
6

Please help me, I don’t know how to solve for x on this one

Mathematics
1 answer:
Brums [2.3K]3 years ago
5 0

Use trigonometry to solve for x.

The tangent function is defined to be opposite side divided by the adjacent side.

tan(78) = x/10

tan(78)(10) = x

47.0463010948 = x

We now round off to two decimal places. In other words, we round off to the nearest tenths.

47.05 = x

Done!

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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
Possible
Alecsey [184]
<h3>Answer:  Independent</h3>

For two events A and B, if the occurrence of either event in no way affects the probability of the occurrence of the other event, then the two events are considered to be <u>  independent  </u> events.

=========================================================

Explanation:

Consider the idea of flipping a coin and rolling a dice. If these actions are separate (i.e. they don't bump into each other), then one object won't affect the other. Hence, one probability won't change the other. We consider these events to be independent.

In contrast, let's say we're pulling out cards from a deck. If we don't put the first card back, then the future probabilities of other cards will change. This is considered dependent.

3 0
2 years ago
Find the least number by which 243 must be divided to make it a perfect<br> square.
yaroslaw [1]

Answer:

Step-by-step explanation:

  • Well I hate to break the news but 243 is not a perfect square. I'll work you through it, 243 is not a perfect square because it is not an even number. an even number must end in (0,2,4,6,8)
  • Step one. Find the square root. the square root of 243 is <em>15.588. </em>
  • Step two Is it a perfect square. No 243 just cant be a perfect square.
  • Hope this helped :)
4 0
3 years ago
A printer prints fewer than 18 pages per minute. What is the maximum number of pages the printer can print in 7 minutes?
max2010maxim [7]

Answer:127 pages of paper

Step-by-step explanation:

18 times 7

4 0
3 years ago
Z= -y + m/x <br><br> Solve for x<br><br><br> Please show work.
Fudgin [204]
Answer is a. You are trying to get x so that would be your answer “x=“ then you have Z and y left. This is still going to be a fraction since there is that line there still and won’t move. M is independent so that will be the denominator. You now have +y-z. This will then be on top and the reason why it’s on top is because most equations have the expression on top so it’s a numerator.
5 0
3 years ago
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