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klasskru [66]
2 years ago
10

Help please ! Find the missing number in the table.

Mathematics
1 answer:
bulgar [2K]2 years ago
4 0
If we take 771 to be the 100%, and increase it by 98%, thus 100% + 98% = 198.

now, if 771 is the 100%, what is the 198%?

\bf \begin{array}{ccll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
771&100\\
a&198
\end{array}\implies \cfrac{771}{a}=\cfrac{100}{198}\implies \cfrac{771\cdot 198}{100}=a
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4.2kg of oranges are £8.76. How much does 2kg cost
seropon [69]

Answer:

x = 4.17

Step-by-step explanation:

We can use a ratio to solve

4.2 k                  2 k

---------------- = ---------------

8.76                     x

Using cross products

4.2 * x = 2 * 8.76

4.2x =17.52

Divide by 4.2

x=4.17

8 0
3 years ago
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Find the lateral area of this square<br> based pyramid.<br> 10 in<br> 5 in<br> [ ? ] in
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The missing answer is 5 in as well
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Mark the points A(5, 5) and B(7.2). Draw a circle that has point A as the center and passes through point B. Measure and label t
olasank [31]

Answer:

i think 4 units im not sure though though

Step-by-step explanation:

4 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
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HELP ASAP- I would give 40 points just for these 3 answers
Slav-nsk [51]
5) The relation between intensity and current appears linear for intensity of 300 or more (current = intensity/10). For intensity of 150, current is less than that linear relation would predict. This seems to support the notion that current will go to zero for zero intensity. Current might even be negative for zero intensity since the line through the points (300, 30) and (150, 10) will have a negative intercept (-10) when current is zero.

Usually, we expect no output from a power-translating device when there is no input, so we expect current = 0 when intensity = 0.


6) We have no reason to believe the linear relation will not continue to hold for values of intensity near those already shown. We expect the current to be 100 for in intensity of 1000.


8) Apparently, times were only measured for 1, 3, 6, 8, and 12 laps. The author of the graph did not want to extrapolate beyond the data collected--a reasonable choice.
7 0
3 years ago
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