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3 years ago

How many solutions are there to the equation 12x+6=5x

Mathematics

1 answer:

Alenkinab [10]3 years ago

4 0

Answer:

One solution; x = - $\frac{6}{7}$

Step-by-step explanation:

12x + 6 = 5x

12x - 5x = -6

7x = -6

x = - $\frac{6}{7}$

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A hat is on sale for h dollars. The regular price is 7 times as much. Charlie has enough money to buy two hats at the regular pr

Nostrana [21]

Answer:

Charlie can buy 14 hats at the sale price

Step-by-step explanation:

Let

h ----> the sale price of the hat

x ---> the regular price of the hat

we know that

$x=7h$ ----> equation A

Two hats at the regular price is equal to multiply 2 by x

Multiply by 2 equation A both sides

$2x=(7h)*2$

$2x=14h$

2 hats at the regular price cost the same that 14 hats at the sale price

therefore

Charlie can buy 14 hats at the sale price

8 0

3 years ago

Jerry was comparing the points the bulls socered for different games he recorded 85,85, 86,77, 93 determine the mean median mode

maks197457 [2]

Answers:
Mean - 85.2
Median - 85
Mode - 85

Explanation:
Mean - 77+85+85+86+93=426/5 (the amount of data points given) = 85.2

Median - Organize the data points from least to greatest and eliminate the outermost numbers two at a time (one from both side) until you’re left with one.

Mode - The number that appears the most often in a given data set

3 0

3 years ago

Which of the following matrices is the solution matrix for the given system of equations? x + 5y = 11 x - y = 5

givi [52]

<h2>The required solution is x = 6 and y = 11 </h2>

Step-by-step explanation:

Given system of equations are

x+5y = 11 and x-y =5

$A= \left[\begin{array}{cc}1&5\\1&-1\end{array}\right]$ $X=\left[\begin{array}{c}x\\y\end{array}\right]$

and $B= \left[\begin{array}{c}11\\5\end{array}\right]$

∴AX=B

$adj A = \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]$

$A= \left|\begin{array}{cc}1&5\\1&-1\end{array}\right|=-6$

∴ $A^{-1} =\frac{adj A}{|A|}$

So, $A^{-1} =\frac{ \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]}{-6}$

$A^{-1} ={ \left[\begin{array}{c \c} {{\frac{1}{6} }}&{\frac{5}{6}}\ \\ {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]}$

$X =A^{-1}\times B$

⇒ $\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c \c} {{\frac{1}{6} }}&{\frac{5}{6}}\ \\ {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]} \times \left[\begin{array}{c}11\\5\end{array}\right]$