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photoshop1234 [79]
3 years ago
12

Working on squared roots​

Mathematics
2 answers:
DerKrebs [107]3 years ago
8 0

Answer:

C or the third one

Step-by-step explanation:

musickatia [10]3 years ago
7 0

Answer:

C: 5\sqrt{3}

Step-by-step explanation:

\sqrt{75} = \sqrt{25} \sqrt{3} =5 \sqrt{3}

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Khan sold his cycle at loss of 8%.had He sold it for 240 more he would have gained 12%
denpristay [2]

Khan sold his cycle at loss of 8%.had He sold it for 240 more he would have gained 12%. Then cost price is $ 1200 and selling price is $ 1104

<h3><u>Solution:</u></h3>

Assume the cost price be "a"

So the selling price of cycle = a - 8% of a

\text { selling price }=a-\frac{8 a}{100}=\frac{23 a}{25}

Now according to given, had He sold it for 240 more he would have gained 12%

Selling price + 240 = cost price + 12% profit

\frac{23a}{25} + 240 = a + \frac{12a}{100}

\frac{23a + 6000}{25} = \frac{100a + 12a}{100}\\\\92a + 24000 = 112a\\\\20a = 24000\\\\a = 1200

So the cost price is $ 1200

Now the selling price = \frac{23a}{25} = \frac{23 \times 1200}{25} = 1104

So the selling price is $ 1104

7 0
4 years ago
I need helpppppppp I don’t like remainders :/
den301095 [7]

Answer:

The correct answer is 67 9/11

4 0
3 years ago
Read 2 more answers
What is 13 - x = -7 what is x in this equation
Zielflug [23.3K]

Answer:

x = 20

(hope i helped)

3 0
3 years ago
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Help me out :) 15 points! Show work, and only answer if you know how to do it correctly.
Volgvan

Answer:

28

Step-by-step explanation:

First u have to find the bisector,then u have to descrese 18

8 0
3 years ago
Solve the system by using a matrix equation.<br> --4x - 5y = -5<br> -6x - 8y = -2
evablogger [386]

Answer:

Solution : (15, - 11)

Step-by-step explanation:

We want to solve this problem using a matrix, so it would be wise to apply Gaussian elimination. Doing so we can start by writing out the matrix of the coefficients, and the solutions ( - 5 and - 2 ) --- ( 1 )

\begin{bmatrix}-4&-5&|&-5\\ -6&-8&|&-2\end{bmatrix}

Now let's begin by canceling the leading coefficient in each row, reaching row echelon form, as we desire --- ( 2 )

Row Echelon Form :

\begin{pmatrix}1\:&\:\cdots \:&\:b\:\\ 0\:&\ddots \:&\:\vdots \\ 0\:&\:0\:&\:1\end{pmatrix}

Step # 1 : Swap the first and second matrix rows,

\begin{pmatrix}-6&-8&-2\\ -4&-5&-5\end{pmatrix}

Step # 2 : Cancel leading coefficient in row 2 through R_2\:\leftarrow \:R_2-\frac{2}{3}\cdot \:R_1,

\begin{pmatrix}-6&-8&-2\\ 0&\frac{1}{3}&-\frac{11}{3}\end{pmatrix}

Now we can continue canceling the leading coefficient in each row, and finally reach the following matrix.

\begin{bmatrix}1&0&|&15\\ 0&1&|&-11\end{bmatrix}

As you can see our solution is x = 15, y = - 11 or (15, - 11).

4 0
3 years ago
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