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1 year ago

Which proportion could be used to find the missing side length?

Mathematics

1 answer:

aliya0001 [1]1 year ago

3 0

Answer:

Step-by-step explanation:

ED : DL = FD : DM

ED / DL = FD / DM

FD = ED·DM / DL

FD = 88 · 84 / 48 = 154

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Sedbober [7]

put in an positive number for x and check

2 - 5 = -3

5 - 2 = 3

- 2 + 5 = 3

5 + (-2) = 2

so the only different one is the first

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3 years ago

Mr. Santon bought 2.6 yards of fabric for $8.94 about how much was the cost per yard

Zanzabum

Since this question is asking for the unit rate of the cost per yard of fabric, think of this as a ratio and proportions question where 2.6:8.94 = 1:x (represents the cost per yard).
To solve, cross multiply to get 2.6x=8.94
Then, solve for x and you get about $3.44 per yard

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3 years ago

A school band consist of 45 boys and 50 girls. Of the total number of boys, 18 play the drums. What percent of the boys in the b

8_murik_8 [283]

That would be 8.1 if you divide

3 0

3 years ago

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If x represents the amount Rita earns each week, which expression represents the amount she earns in a year?

Bezzdna [24]

52x because there are approximately 52 weeks in a year.

6 0

2 years ago

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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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