All categories

2 years ago

Solve for x: 2/7 (x-2) =4c

Mathematics

1 answer:

shutvik [7]2 years ago

8 0

Answer:

x=14c+2

Step-by-step explanation:

$\frac{2}{7} (x-2)=4c \\\frac{2}{7}x-\frac{4}{7} =4c\\\frac{2}{7}x=4c+\frac{4}{7}\\\frac{2}{7}x= \frac{28c+4}{7} \\2x=28c+4\\x=14c+2$

You might be interested in

3/8 + 1/4 + 1/2 - 2/3 =

Nostrana [21]

Answer:

$\frac{11}{24}$

Step-by-step explanation:

3/8 + 1/4 + 1/2 - 2/3

- > 1/4 = 2/8

3/8 + 2/8 + 1/2 - 2/3

5/8 + 1/2 - 2/3

- > 1/2 = 4/8

5/8 + 4/8 - 2/3

9/8 - 2/3

- > LCM of 8,3: 24

- > 9/8 = 27/24

- > 2/3 = 16/24

27/24 - 16/24

11/24

Hope this helps you.

7 0

3 years ago

If sin =1/2then show that 3cos -4cos^3=0z

Katarina [22]

Step-by-step explanation:

ever coner should have a value.... for example sinA or cos$

please check the question again

7 0

3 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:

$\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|$

$\Delta t_{1/2} \approx 65.788\,days$

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

4 0

3 years ago

Read 2 more answers

Rewrite in simplest terms:

tangare [24]

Answer:

If one −5s−7(8s−1): -61s+7

If two −5s−7(8s−1): -112s+14

Step-by-step explanation:

-5s-7(8s-1)

Multiply -7 onto 8s and -1:

-5s-56s+7

add -5s and -56s:

-61s+7

−5s−7(8s−1)−5s−7(8s−1)

Multiply both -7 to 8s and -1

-5s-56s+7-5s-56s+7

add:

-112s+14

6 0

3 years ago

What two numbers multiply to get 40 and add to get 11

Trava [24]

I got the factors of 40 but both numbers do not add up to 11. Negative numbers are not considered because the product and sum are positive numbers.

1 x 40 = 40 1 + 40 = 41
2 x 20 = 40 2 + 20 = 22
4 x 10 = 40 4 + 10 = 14
5 x 8 = 40 5 + 8 = 13

I also got the addends of 11 and solved each set to get the product
1 + 10 = 11 1 x 10 = 10
2 + 9 = 11 2 x 9 = 18
3 + 8 = 11 3 x 8 = 24
4 + 7 = 11 4 x 7 = 28
5 + 6 = 11 5 x 6 = 30

6 0

3 years ago