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3 years ago

What is 36 as a fraction of 60 in its simplest form?

1 answer:

3 0

Think: "what's the greatest common factor of both 36 and 60?"
12
Divide both #s by 12 and put them in fraction form.

36 ÷ 12 = 3
60 ÷ 12 = 5

3/5 is your final answer.

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Match the mathematical expression with its translation

egoroff_w [7]

Answer:

$\begin{array}{cc}\text{Mathematical expression}&\text{Translation}\\ \\xy&\text{The product of two numbers}\\ \\\dfrac{x}{y}&x\text{ divided by }y\\ \\x-y&x\text{ minus }y\\ \\x+y&\text{the sum of }x\text{ and }y\\ \\y-x&x\text{ subtracted from }y\\ \\y:x&y\text{ divided by } x\\ \\x+y=6&\text{the sum of two numbers is 6}\\ \\xy=6&\text{the product of two numbers is 6}\\ \\6x=y&\text{6 times a number equals }y\\ \\y=x-6&\text{6 less than a number is }y\end{array}$

Step-by-step explanation:

6 0

3 years ago

PLEASE HELP!!!!!!!! IMAGE ATTACHED!!!!

erma4kov [3.2K]

Answer:

Step-by-step explanation:

4) (-2,3) ; (-1,-2)

$Slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\=\frac{-2-3}{-1-[-2]}\\\\=\frac{-5}{-1+2}\\\\=\frac{-5}{1}\\\\=-5$

5) line is parallel to x-axis. So, slope= 0

6) (1,1) ; (-2, -1)

$Slope=\frac{-1-1}{-2-1}\\\\=\frac{-2}{-3}\\\\=\frac{2}{3}$

6 0

3 years ago

Y+2=2/5(x+6) write in standard form

icang [17]

y+2=(2/5)(x+6)

5(y+2)=2(x+6)

5y+10=2x+12

-2x+5y+10-12=0

-2x+5y-2=0

2x-5y+2=0

8 0

3 years ago

Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,

sweet [91]

Answer:

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

$f'' = x - \frac{3}{2}$

$f' = \int {\left(x-\frac{3}{2}\right) } \, dx$

$f' = \int {x} \, dx -\frac{3}{2}\int \, dx$

$f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C$ , where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative ( $f'(4) = 1$ ):

$1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C$

$C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)$

$C = -1$

The first derivative is $y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1$ , and the particular solution is found by integrating one more time and using the initial condition ( $f(0) = 0$ ):