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Alexeev081 [22]
2 years ago
14

2 more than the difference between 8 and h

Mathematics
2 answers:
mylen [45]2 years ago
8 0

Answer:

2+8-h

Step-by-step explanation:

hope this helps :)

Harman [31]2 years ago
3 0
8-h+2

hope this helped! :)
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Find the next two<br> numbers in the<br> sequence. What is<br> the rule?<br> 100, 25, 50, 12:5, 25,
Tema [17]

Answer:

6.25, 12.5

Step-by-step explanation:

The sequence goes divide by 4 then multiply by 2

6 0
2 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
Can someone help with this please
nasty-shy [4]

Answer:

22x^2 -10xy + 12x

Step-by-step explanation:

To add and subtract polynomials, combine only like terms. When doing it vertically, stack polynomials and line up the same bases and exponents. Once this is done, simply add and subtract the coefficients.

  10x^2 + 12xy + 4x

+ 12x^2 -22xy + 8x

_________________

22x^2 -10xy + 12x

Here we added 10 + 12 = 22, 12 + -22 = -10 and 4 + 8 = 12.

6 0
3 years ago
Jem boy wants to make his 8-meter square pool into a rectangular one by increasing its length by 2m and decreasing its width by
butalik [34]
Side = 8m
He increases length by 2m and reduces width by 2m
(a.) New dimensions will be -
Length = 10m and Width = 6m
(b.) Area = length x breadth (or width)
 here we will use the special product (x+y) (x-y)
length = (x+y) and width = (x-y)
area = (8+2) x (8-2)
            8^(2) - 2^(2) 
            = 60
3 0
2 years ago
The girls and boys' basketball teams sponsored a car wash that made $109. There were twice as many girls as
V125BC [204]

Answer:

Boys - 36$ 33 cents

Girls - 72$ 67 cents

Step-by-step explanation:

Boys x

Girls 2x

Total 109

2x+x=109

3x=109/:3

X=36,3333

Boys x =36$ 33c

Girls 2x= 72,666= 72$ 67c

7 0
3 years ago
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