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erik [133]
3 years ago
8

Which graph shows the original figure rotated 180 degrees

Mathematics
2 answers:
Serjik [45]3 years ago
8 0

Answer:

I believe it would be B.

Step-by-step explanation:

Half a circle is 180 degrees, which is the location of option B. In other words you just rotate it halfway around and you'll find it at 180°.

Sphinxa [80]3 years ago
7 0
It should be B because 180° should transfer it to the opposite quadrant and change the signs of both x and y
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The lengths of the sides of a triangle are given below. Classify the triangle as acute, obtuse, or right.
enot [183]

Answer:

  • B. Obtuse Triangle

Step-by-step explanation:

<u>Use Pythagorean to identify if the triangle is right:</u>

  • 9² = 7² + 3²
  • 81 = 58

We see the triangle is <u>not </u>right and the difference of the squares tells us the angle opposite to side with measure of 9 is greater than 90°.

It means the triangle is <u>obtuse</u>.

Correct choice is B

8 0
3 years ago
What is the slope of the line?<br><br> ________
kondor19780726 [428]

Answer:

3 over 1

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5 0
4 years ago
PLEASE HELP WORTH 16 POINTS WILL REPORT LINKS
Igoryamba

Answer:

Angle A = 30.8

Angle B = 33.2

Step-by-step explanation:

180 - 116 = 64  

This means that the other two angles need to add up to 64.

I just plugged in numbers until it worked.

I got that x = 14.6

2x + 4

2(14.6) + 4

29.2 + 4

33.2

3x - 13

3(14.6) - 13

43.8 - 13

30.8

Those are your angles. To make sure they are correct you just need to add up all the angles and make sure that they equal to 180.

116 + 33.2 + 30.8 =

180

6 0
3 years ago
Read 2 more answers
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
A person burn 55 calories an hour while they are sleeping. How many calories does a person burn during a 15 minute nap?
BaLLatris [955]

Answer:

13.75

OR

14

Step-by-step explanation:

55 calories are burned an hour while sleeping

There's 60 minutes in an hour

55/60 = 0.916666667

there's 0.916666667 calories burned in a minute

13.75 calories are burnt in 15 minutes.

If you round 13.75, it becomes 14.

7 0
4 years ago
Read 2 more answers
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