A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?

Which method correctly solves the equation using the multiplication property of equality and the reciprocal of One-third? One-th

Airida [17]

Answer:

Step-by-step explanation:

Given the equation $\frac{1}{3}(x+9) = -12$

Step 1;

Expand the bracket at the right hand side of the equation to have:

$\frac{1}{3}x +\frac{1}{3}(9) = -12\\\frac{1}{3}x+3=-12\\subtracting\ 3\ from\ both\ sides\\\frac{1}{3}x+3-3=-12-3\\\frac{1}{3}x=-15\\$

Taking the reciprocal of both sides:

$\frac{3}{x} = \frac{-1}{15} \\ cross\ multiplying\\-x =45\\x=-45$

4 0

2 years ago

Read 2 more answers

16 > - 8 + x A. x > 24 B. x < 24 C. x > 8 D. x < 8

xeze [42]

Answer:

B. x < 24

Step-by-step explanation:

16 > - 8 + x

+8 > +8 + x Add 8 to both sides

24 > x

Final answer: x < 24

6 0

2 years ago

-2(3x + 4) + 7x = 10

musickatia [10]

Answer:

18

Step-by-step explanation:

-2(3x+4)+7x=10

-6x-8+7x=10

x=10+8

x=18

6 0

2 years ago

Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot

Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

$\frac{dm}{dt} = -\frac{t}{\tau}$

Where:

$m$ - Current mass of the isotope, measured in kilograms.

$t$ - Time, measured in days.

$\tau$ - Time constant, measured in days.

The solution of the differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

$\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}$

$\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }$

The half-life of a isotope ( $t_{1/2}$ ) as a function of time constant is:

$t_{1/2} = \tau \cdot \ln2$

$t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2$

The half-life difference between isotope B and isotope A is:

$\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|$

If $\frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9$ , $t_{A} = 33\,days$ and $t_{B} = 43\,days$ , the difference in the half-lives of the isotopes is: