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3 years ago

Find the equation of a line that runs through (2,5) and is parallel to the line with a slope of 2/4

1 answer:

6 0

Parallel lines have the same slope

y - y1 = m(x - x1)
slope(m) = 2/4
(2,5)...x1 = 2 and y1 = 5
now we sub
y - 5 = 2/4(x - 2)

now if there is a typo and ur slope is 3/4....it would be :
y - 5 = 3/4(x - 2)

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The difference between number 20 and its opposite is what percentage of 200? (if you do not remember: opposite of 15 is −15; opp

Lesechka [4]

The opposite of 20 is -20 and the difference is 40. If we put 40/200 we get 0.2, or 20%.

4 0

3 years ago

Read 3 more answers

Radioactive Decay

SpyIntel [72]

Answer:

Percentage of (226Ra) after 900 years is 68%

Step-by-step explanation:

Let P(t) be the amount of (226Ra) present at any time t

Half life of (226 Ra) = 1599 years

If P₀ is initial amount of (226 Ra) then after 1599 years

P(1599)=P₀/2

Decay i amount of radioactive substance is related to time t as

$\frac{dP}{dt}=kP(t)\\\\\frac{1}{P}\,dP=kdt\\\\Integrating\,\, both\,\,sides\\\\ln|P|=kt+c\\\\P(t)=Ce^{kt}\\\\at \,\, t=0\,\, P(0)=P_{o}\\\\P(0)=Ce^{k0}\\\\P_{o}=C\\\\then\\\\P(t)=P_{o}e^{kt}$

To find value of k

$at\,\, t=1599\,years\\\\P(1599)=\frac{P_{o}}{2}\\\\then\\\\\frac{P_{o}}{2} =P_{o}e^{k(1599)}\\\\\frac{1}{2} =e^{k(1599)}\\\\ln|\frac{1}{2}|=k(1599)\\\\k=\frac{ln|\frac{1}{2}|}{1599}=-4.3\times 10^{-3}\\\\\implies P(t)=P_{o}e^{-4.3\times 10^{-3}t}\\\\at\,\, t=900 \\\\P(900)=P_{o}e^{-4.3\times 10^{-3}(900)}\\\\P(900)=0.68P_{o}$

Percentage of radioactive element is:

Amount after 900 years $=\frac{P(900)}{P_{o}}\times 100\\\\=\frac{0.68P_{o}}{P_{o}}\times 100\%\\\\=68\%$

3 0

3 years ago

A circle has a radius of 9 inches. The Radius is multiplied by 2/3 to form a second circle. How is the ratio of the areas relate

liraira [26]

Answer:

$\frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)} = \frac{81}{36} = (\frac{r_{1} }{r_{2}}) ^{2}$

The above expression shows that ratios of the areas of the circles are equal to the square of the ratio of their radii.

Step-by-step explanation:

Radius of first circle $(r_{1})$ = 9 inches

Area of first circle = $\pi r_{1} ^{2}$

Area of first circle = 9 × 9 × π = 81 π

Now, since the radius is multiplied by 2/3 for from a new circle.

∴ Radius of the second circle = $9 \times \frac{2}{3} = 6\ inches$

Area of second circle = $\pi r_{2} ^{2}$

Area of second circle = 6 × 6 × π = 36 π

Now,

$\frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)} = \frac{81\pi }{36\pi }$

$\frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)} = \frac{81}{36} = (\frac{9}{6}) ^{2} = (\frac{r_{1} }{r_{2}}) ^{2}$

∵ $(r_{1})$ = 9 inches and $(r_{2})$ = 6 inches

The above expression shows that ratios of the areas of the circles are equal to the square of the ratio of their radii. i.e., $\frac {radius\ of\ first\ circle)^{2} }{(radius\ of\ second\ circle)^{2} } = \frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)}$

8 0

3 years ago

Omar has 2 3/4 cups of dough to make Dumplings. If he uses 3/16 cup of dough for each dumpling,how many whole dumplings can Omar

Brrunno [24]

Omar can make 14 whole dumplings. You change 2 3/4 cups to an improper fraction ( multiply the denominator and add the numerator ) and you get 11/4. Since you need 3/16 cups for one dumpling you have to find the greatest common denominator which is 16. You multiply 4 by 11 to get the numerator and end up with 44/16. Since you need 3/16 for one whole dumping you divide 44 by 3 and get 14.6 repeating. You cannot have a fraction of a dumpling so you round down and get the answer 14.

5 0

4 years ago

Plssss help

Andre45 [30]

Answer:

A) $\displaystyle \frac{1}{77}$ .
B) $\displaystyle \frac{12}{77}$ .
C) $\displaystyle \frac{4}{11}$ .

Step-by-step explanation:

All marbles here are identical. Also, the question isn't concerned about the order in which the marbles are drawn. Thus, all calculations here shall be combinations rather than permutations.

How many ways to choose three out of six identical red marbles without replacement?

$\displaystyle _6C_3 = c(6, 3) = {6\choose 3} = 20$ .

Note that these three expressions are equivalent. They all represent the number of ways to choose 3 out of 6 identical items without replacement.

How many ways to choose three out of all the 6 + 10 + 6 = 22 marbles?

$\displaystyle _{22} C_{3} = 1540$ .

The probability of choosing three red marbles out of these 22 marbles will be:

$\displaystyle \frac{\text{Number of ways for choosing three out of six red marbles}}{\text{Number of ways to choose three out of 22 marbles}} = \frac{20}{1540} = \frac{1}{77}$ .

How many ways to choose two out of six identical red marbles without replacement?

$\displaystyle _6 C_2 = 15$ .

How many ways to choose one out of 10 + 6 = 16 non-red marbles?

$_{16} C_1=16$ .

Choosing two red marbles does not influence the number of ways of choosing a non-red marble. Both event happen and are independent of each other. Apply the product rule to find the number of ways of choosing two red marbles and one non-red marble out of the pile of 22.

$_6 C_2 \cdot _{16} C_1= 240$ .

Probability:

$\displaystyle \frac{240}{1540} = \frac{12}{77}$ .

Double check that the order doesn't matter here.

None of the marbles are red. In other words, all three marbles are chosen out of a pile of 10 + 6 = 16 white and blue marbles. Number of ways to do so:

$_{16} C_{3} = 560$ .

Probability:

$\displaystyle \frac{560}{1540}= \frac{4}{11}$ .

5 0

4 years ago

Read 2 more answers