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KATRIN_1 [288]
3 years ago
5

Draw a line from each rectangular prism to its matching group of face shapes

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
7 0

Answer:

1 goes with the second face shapes option, 2 goes with the third and 3 goes with the first.

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I don’t understand this problem
ikadub [295]

Hello! the answer is c

First of all let's eliminate the answers that do not make sense. For A. numbers are NOT irrational because they are a fraction. Numbers are also not irrational because they are a repeating number. C and D are both correct as fractions and terminating decimals are both rational numbers.

34/3≈11.33333333

By using a certain formula you can convert repeating decimals to a fraction. But most people know that .3333333 is 1/3. This gives us 11 1/3 as our answer. So what does it match? It is not a terminating decimal. It repeats but can be written as a fraction. Therefore our answer is C) Rational, because it is a fraction.

I hope this helps!

bran-list please

4 0
3 years ago
Lena is asked to write an explicit formula for the graphed geometric sequence (3,6.25) (2,2.5) (1,1)
harkovskaia [24]
The correct answer for this question is this one: "2.5"
Lena is asked to write an explicit formula for the graphed geometric sequence. The value, written as a decimal, should Lena use as the common ratio that is 2.5There exists the same question from other source with the image is shown.
Hope this helps answer your question and have a nice day ahead.
3 0
3 years ago
Read 2 more answers
Find the y-intercept of the line on the graph.
BARSIC [14]
It's a little hard to see, but I'm almost positive the y intercept is -1.
 
Remember, the y intercept is wherever the line intercepts in the y axis 0.

Good luck my man.
5 0
2 years ago
Read 2 more answers
Work out the number of sides tile A has.<br> Both tile A and tile B are regular polygons
finlep [7]

Answer:

12

Step-by-step explanation:

since tile B is a regular polygon means it is an equilateral triangle. The angle inside B would be  60.

(360 - 60)/2 = 150 (interior angle of tile A)

180 - 150 = 30 (exterior angle of tile A)

therefore number of sides = 360 /30 = 12

8 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
2 years ago
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