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anyanavicka [17]
3 years ago
8

1. match the graph to the table of values

Mathematics
1 answer:
Verdich [7]3 years ago
6 0

Answer:

In order (Problem 1-4)

B

C

A

D

Step-by-step explanation:

These will give you 100%! Wish I had these before I took the quick check.

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Calculate the product using partial products:
Archy [21]
A. 400*45.2=400*45+400*0.2=18000+80=18080
B. 14.9*100=15*100-0.1*100=1500-1=1490
C. 76.2*200=76*200+0.2*200=15200+4=15204
5 0
3 years ago
Write the standard form of the equation of each line given the slope and y-intereept.
Elena-2011 [213]

Answer:

5y + 15x = 25, 9x - y = -4

Step-by-step explanation:

The standard form is Ax + By = C. We can make this simple by first creating the slope intercept equations of both questions:

1.  y = (-3/5)x + 5

2. y = 9x + 4

For the first one, we must first get rid of the denominator by multiplying everything by 5, so we get 5y = -15x + 25, and then you can add 3x to both sides to arrive at your first standard eqn, 5y + 15x = 25.

For the second, all you need to do is swap the y and the 4, subtracting both y and 4 from each side, arriving at 9x - y = -4

6 0
3 years ago
Make a conjecture about the diagram below. Do you think you can conclude that △JKL ≅ △XYZ? Explain your reasoning.
konstantin123 [22]

Answer:

Yes, you can conclude that ΔJKL ≅ XYZ.

Step-by-step explanation:

I forget how to describe it in geometric terms but, if you read the similarity statement again, and follow each angle, you'll notice that the shapes match, and they correspond to each other, it would be different if it said something like ΔLKJ ≅ XYZ because it does not match.

If you're looking for a specific reason, we'd say that they're congruent because of SAS (Side-Angle-Side)

If this still confuses you, hit me up.

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
What does x equal, 55°+54°+×+74
lakkis [162]
SOLUTION:

= 55° + 54° + x + 74°

= x + 183°

There is no answer to this expression.
4 0
3 years ago
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