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Verdich [7]
2 years ago
9

Find the surface area of the right rectangular prism shown below. units^2

Mathematics
2 answers:
ziro4ka [17]2 years ago
4 0

Step-by-step explanation:

Formula = A=2(wl+hl+hw)

  • L = 4
  • W = 2
  • H = 2.5

=> A=2(wl+hl+hw)

=> A = 2((2)(4)+(2.5)(4) + (2.5)(2))

=> A = 2(8 + 10 + 5)

=> A = 2(23)

=> A = 46

\therefore Surface area = 46

Semmy [17]2 years ago
4 0

Answer:

A=46

Step-by-step explanation:

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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
Jo's collection contains US, Indian and British stamps. If the ratio of US to Indian stamps is 5 to 2 and the ratio of Indian to
andreev551 [17]

Answer:(E). 25:2

Step-by-step explanation:

First ratio:

Us : India is 5 : 2

Second ratio

India : British is 5 : 1

multiply(5:2) by 5 and multiply (5:1) by 2 to balance the India stamps.

Therefore now

Us : India is 25 : 10

India : British is 10 : 2

The ratio of us : British is 25:2

5 0
3 years ago
Which line corresponds to y=2x+5? <br><br> *please see picture*<br> *serious answers only*
Ivahew [28]

Answer:

B

Step-by-step explanation:

the one on the tight becasue try plugging in zero

8 0
3 years ago
Read 2 more answers
HELP ME PLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS
WARRIOR [948]

Answer:

volume = 240 in.^2

240 times 1/2 = 240/2

240/2 = 120

Step-by-step explanation:

4 0
3 years ago
For what values of h and k does the linear system have infinitely many solutions?
Sophie [7]

Answer:

h=-32 and k=-24

Step-by-step explanation:

4x_1 + 3x_2 = -1:a_1x_1 + b_1x_2=c_1\\hx_1 +kx_2=8:a_2x_1+b_2x_2=c_2

For infinitely many solutions, lines must be same;\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\\\=\frac{4}{h}=\frac{3}{k}=\frac{-1}{8}\\\\h=-32, k=-24

7 0
3 years ago
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