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3 years ago

PLS HELP ME ASAP FOR 15!! (MUST SHOW WORK!!) + LOTS OF POINTS!!

Mathematics

1 answer:

DIA [1.3K]3 years ago

4 0

If the area of the old garden was 1m^2, then the side lengths were both 1
if she increases the lengths by 50%, then 1 turns into 1.5
1.5 times 1.5 is 2.25
The area of her new garden is 2.25m^2

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Is 10s equal one thousands

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Ten times ten equals one hundred, if that's what you are asking for.

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3 years ago

Find the unit rate. 16 miles in 4 hours

MatroZZZ [7]

Answer:

4 miles per hour

Step-by-step explanation:

divide 16 by 4

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4 years ago

Ethan bought 3 pints of raspberries for $5 per pint and x pints of blueberries for $3 per pint. The average

bagirrra123 [75]

The y-intercept represents the average cost, in dollars per pint, of all the raspberries Ethan bought. Therefore, option B is the correct answer.

Given that, Ethan bought 3 pints of raspberries for $5 per pint and x pints of blueberries for $3 per pint.

The average is (15 + 3x)/(3+x) per pint.

We need to find what quantity will be represented by the y-intercept of the graph.

<h3>How to find the y-intercept of the graph?</h3>

Since the y-intercept always has a corresponding x-value of 0, replace x with 0 in the equation and solve for y. On a graph, the y-intercept can be found by finding the value of y when x=0. This is the point at which the graph crosses through the y-axis.

Now, the graph of y=(15 + 3x)/(3+x)is given below.

To find the x-intercept, substitute in 0 for y and solve for x. To find the y-intercept, substitute in 0 for x and solve for y.

x-intercept(s): (−5,0)

y-intercept(s): (0,5)

So, the y-intercept represents the average cost, in dollars per pint, of all the raspberries Ethan bought.

Therefore, option B is the correct answer.

To learn more about y-intercept visit:

brainly.com/question/14180189.

#SPJ1

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1 year ago

Can -2/5 be simplified

sergiy2304 [10]

No it can't because nothing will go into 2 or 5 at the same time...

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4 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

3 years ago