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1 year ago

Please help me it is easy I just dont understand

Mathematics

1 answer:

postnew [5]1 year ago

4 0

Answer:

13%

15%

41%

31%

Step-by-step explanation:

To find the percentage one amount is out of another amount, divide the smaller amount by the bigger amount. You will end up with a decimal (0.xy where x is the 10% place and y is the 1% place).

For example: to find the percentage that 39 is out of 300 students, we divide 39 by 300. This gives us the decimal 0.13. Written as a percentage, that is 13%.

We can do this for all of them.

Here is the math for each one:

39 ÷ 300 = 0.13 → 13%

45 ÷ 300 = 0.15 → 15%

123 ÷ 300 = 0.41 → 41%

93 ÷ 300 = 0.31 → 31%

Once you have all these decimals, you might choose to check your work by adding them up. They should at up to 1.00, or 100%. In this case, they do, so we know these answers are correct.

I hope this helps. Please let me know if you have questions.

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Solving a Two-Step Matrix Equation<br> Solve the equation:

Cloud [144]

Answer:

$\boxed {x_{1} = 3}$

$\boxed {x_{2} = -4}$

Step-by-step explanation:

Solve the following equation:

$\left[\begin{array}{ccc}3&2\\5&5\\\end{array}\right] \left[\begin{array}{ccc}x_{1}\\x_{2}\\\end{array}\right] + \left[\begin{array}{ccc}1\\2\\\end{array}\right] = \left[\begin{array}{ccc}2\\-3\\\end{array}\right]$

-In order to solve a pair of equations by using substitution, you first need to solve one of the equations for one of variables and then you would substitute the result for that variable in the other equation:

-First equation:

$3x_{1} + 2x_{2} + 1 = 2$

-Second equation:

$5x_{1} + 5x_{2} + 2 = -3$

-Choose one of the two following equations, which I choose the first one, then you solve for $x_{1}$ by isolating

$3x_{1} + 2x_{2} + 1 = 2$

-Subtract $1$ to both sides:

$3x_{1} + 2x_{2} + 1 - 1 = 2 - 1$

$3x_{1} + 2x_{2} = 1$

-Subtract $2x_{2}$ to both sides:

$3x_{1} + 2x_{2} - 2x_{2} = -2x_{2} + 1$

$3x_{1} = -2x_{2} + 1$

-Divide both sides by $3$ :

$3x_{1} = -2x_{2} + 1$

$x_{1} = \frac{1}{3} (-2x_{2} + 1)$

-Multiply $-2x_{2} + 1$ by $\frac{1}{3}$ :

$x_{1} = \frac{1}{3} (-2x_{2} + 1)$

$x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$

-Substitute $-\frac{2x_{2} + 1}{3}$ for $x_{1}$ in the second equation, which is $5x_{1} + 5x_{2} + 2 = -3$ :

$5x_{1} + 5x_{2} + 2 = -3$

$5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3$

Multiply $-\frac{2x_{2} + 1}{3}$ by $5$ :

$5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3$

$-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3$

-Combine like terms:

$-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3$

$\frac{5}{3}x_{2} + \frac{11}{3} = -3$

-Subtract $\frac{11}{3}$ to both sides:

$\frac{5}{3}x_{2} + \frac{11}{3} - \frac{11}{3} = -3 - \frac{11}{3}$

$\frac{5}{3}x_{2} = -\frac{20}{3}$

-Multiply both sides by $\frac{5}{3}$ :

$\frac{\frac{5}{3}x_{2}}{\frac{5}{3}} = \frac{-\frac{20}{3}}{\frac{5}{3}}$

$\boxed {x_{2} = -4}$

-After you have the value of $x_2$ , substitute for $x_{2}$ onto this equation, which is $x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$ :

$x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$

$x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}$

-Multiply $-\frac{2}{3}$ and $-4$ :

$x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}$

$x_{1} = \frac{8 + 1}{3}$

-Since both $\frac{1}{3}$ and $\frac{8}{3}$ have the same denominator, then add the numerators together. Also, after you have added both numerators together, reduce the fraction to the lowest term:

$x_{1} = \frac{8 + 1}{3}$

$x_{1} = \frac{9}{3}$

$\boxed {x_{1} = 3}$

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2 years ago