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3 years ago

in a rectangle, the ratio of the width to the length is 4:5. If the rectangle is 40 centimeters long find its width

Mathematics

1 answer:

dybincka [34]3 years ago

4 0

Gofipanswer.com copy and paste ur answer it showes the answer and explains it good luck god bless<3

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A1 Plumbing Service charges $35 per hour plus a $25 travel charge for a service call. Good Guys Plumbing Repair charges $40 per

AlladinOne [14]

A1 = 35x+25

Good guys = 40x

35x+25 = 40x

subtract 3x x from each side

25 = 5x

divide each side by 5

x = 25/5 = 5

so 5 hours they would be the same cost, anything over 5 hours and A1 would be better

so X>5 for A1 to be the better price

3 0

2 years ago

I did 4 more problems than Manuel how many problems did I do

Aleksandr-060686 [28]

4 more than Manuel did

5 0

3 years ago

HELP URGENT GET IT CORRECT ALL OF THEM

Nikolay [14]

Answer:

2/9

Step-by-step explanation:

1/3 * 2/3 = 2/9

Answer:

136/9

Step-by-step explanation:

4 1/4 * 2 2/3 = 17/3 * 8/3 = 136/9 = 15 1/9

6 0

3 years ago

Write an ex expression for the product of 11 and w

klemol [59]

Answer:

11w

Step-by-step explanation:

product means multiply

6 0

3 years ago

Plz help<br>urgent !!!!<br>will give the brainliest !!

Mkey [24]

Answer:

$X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}$

Step-by-step explanation:

The question we have at hand is, in other words,

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix}$ - where we have to solve for the value of $X$

If we have to isolate $X$ here, then we would have to take the inverse of the following matrix ...

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}$ ... so that it should be as follows ... $\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}$

Therefore, we can conclude that the equation as to solve for " $X$ " will be the following,

$X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ - First find the 2 x 2 matrix inverse of the first portion,

$\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}$ = $\frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}$ = $\frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix}$ = $\begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}$

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

$X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ ,

$X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix}$ - Simplifying this, we should get ...

$\begin{bmatrix}1&3\\ 2&4\end{bmatrix}$ ... which is our solution.

7 0

3 years ago