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valina [46]
3 years ago
14

the rectangle below has a perimeter of 120 in right an equation to solve for x and then calculate the length and width of the re

ctangle.​

Mathematics
2 answers:
denis-greek [22]3 years ago
8 0

Answer:

<h2>Given</h2>
  1. <u>Length</u><u> </u><u>=</u><u> </u><u>x </u><u>+</u><u>7</u>
  2. <u>Bredth</u><u> </u><u>=</u><u> </u><u>3</u><u>x</u><u>-</u><u>3</u>
  3. <u>perimeter</u><u>=</u><u>1</u><u>2</u><u>0</u><u> </u><u>inches</u>
<h2><u>To </u><u>find</u></h2>
  • <u>The</u><u> </u><u>Value</u><u> </u><u>of </u><u>X</u>
<h2><u>Solution</u></h2>
  • <u>we </u><u>know</u><u> that</u><u> </u><u>the </u><u>perimeter</u><u> </u><u>of </u><u>a </u><u>rectangle</u><u> </u><u>=</u><u> </u><u>2</u><u>(</u><u>Length</u><u>+</u><u>Bredth)</u>
  • <u>Inserting</u><u> </u><u>the </u><u>values</u>
  • <u>=</u><u>></u><u>1</u><u>2</u><u>0</u><u> </u><u>inches </u><u>=</u><u> </u><u>2</u><u>(</u><u>x </u><u>+</u><u>7</u><u>+</u><u>3</u><u>x</u><u> </u><u>-</u><u>3</u><u>)</u>
  • <u>=</u><u>></u><u>1</u><u>2</u><u>0</u><u> </u><u>inches </u><u>=</u><u>2</u><u>(</u><u>4</u><u>x</u><u>+</u><u>4</u><u>)</u>
  • <u>=</u><u>></u><u>1</u><u>2</u><u>0</u><u>i</u><u>n</u><u>c</u><u>h</u><u>e</u><u>s</u><u>/</u><u>2</u><u>=</u><u> </u><u>4</u><u>x</u><u>+</u><u>4</u>
  • <u>=</u><u>></u><u>6</u><u>0</u><u> </u><u>inches </u><u>=</u><u>4</u><u>x</u><u>+</u><u>4</u>
  • <u>=</u><u>></u><u>6</u><u>0</u><u>i</u><u>n</u><u>c</u><u>h</u><u>e</u><u>s</u><u> </u><u>-</u><u>4</u><u> </u><u>=</u><u> </u><u>4</u><u>x</u>
  • <u>=</u><u>></u><u>4</u><u>x</u><u> </u><u>=</u><u> </u><u>5</u><u>6</u><u> </u><u>inches</u>
  • <u>=</u><u>></u><u>x </u><u>=</u><u> </u><u>5</u><u>6</u><u>i</u><u>n</u><u>c</u><u>h</u><u>e</u><u>s</u><u> </u><u>/</u><u>4</u>
<h2><u>=</u><u>></u><u>x </u><u>=</u><u> </u><u>1</u><u>4</u></h2>
Korolek [52]3 years ago
7 0

Answer:

I did what I can

Hope this helps :)

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Question Which of these strategies would eliminate a variable in the system of equations? 2x - y = 11 5x + 3y = -11 How many pos
kenny6666 [7]

Answer:

Elimination (multiplication and subtraction)

There is 1 solution

Step-by-step explanation:

The only way to solve this system of equations is by elimination (multiplication). Let's say that we want to cancel out the x. You multiply the first equation by 5 and you multiply the second equation by 2.

Your equations now are:

10x-5y=55 and 10x+6y=-22

You can subtract these equations to get -11y = 77.

Now that we know that y = -7, we can substitute y in to solve x.

10x-5(-7) = 55

10x=20

x=2

Final answer (2,-7)

3 0
3 years ago
A rectangle has an area of 1/6 square centimeters and a length of 1.5 centimeters. what is the width? what is the perimeter?
SCORPION-xisa [38]
The area of a rectangle can be calculated by the product of its length and the width. Here we are given the area and the length so we can simply calculate for width. The perimeter is the sum of all the side lengths of the shape. We calculate as follows:

Area = lxw
1/6 = 1.5w
w = 1/9

Perimeter = 2l + 2w
Perimeter = 5/9

Hope this answers the question. Have a nice day.
6 0
3 years ago
What is the least common denominator of 5, 3, 4, 6?????..!!! (PLEASE ANSWER THIS QUESTION) I AM IN A BIT HURRY!!
sergij07 [2.7K]

Answer:

60

skip count to find lest common denominator quickly



4 0
3 years ago
Ellie drew ΔLMN, in which m∠LMN = 90°. She then drew ΔPQR, which was a dilation of ΔLMN by a scale factor of one half from the c
vaieri [72.5K]

Answer:

m∠P ≅ m∠L; this can be confirmed by translating point P to point L.

Step-by-step explanation:

Angle angle (AA) similarity postulate state that two triangles are similar if two of their corresponding angle is similar. The corresponding angle for each point of the triangles will be:

∠L=∠P

∠Q=∠M

∠N=∠R

Since the 2nd triangle made from dilation, it should maintain its orientation.

Option 1 is true, ∠P corresponds to ∠L. If you move/translate point P to point L, you can confirm it because their orientation is the same.

Option 2 is false, the triangle will be similar if ∠P=∠N but you can't confirm it with translation alone.

Option 3 and 4 definitely wrong because it speaking about length, not the angle.

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
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