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ra1l [238]
3 years ago
10

There are 5 green jelly beans, 7 red jelly beans and 8 yellow jelly beans. What percent of the jelly beans are red?

Mathematics
1 answer:
s2008m [1.1K]3 years ago
5 0

Answer: 35%

Step-by-step explanation:

8 + 5 + 7 = 20

50% = 10

25% = 5

10% = 2

5% = 1

25% + 10% = 5+2

35% = 7

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One - eighth of 7.92 x 10^10
liraira [26]
The answer is 9900000000 (thats 8 zeros) What you do is you find 10^10 which is 10000000000 (ten zeros) and multiply that by 7.92 to get 79200000000 then you just multiply it by 1/8 to get the answer.
3 0
3 years ago
Scientific notation for 3427
Deffense [45]

Answer:

3.427 *10^3

Step-by-step explanation:

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the 10 . If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.

5 0
3 years ago
A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the
ohaa [14]

Answer:

a) t = 2 *ln(\frac{82}{5}) =5.595

b) t = 2 *ln(-\frac{820}{p_0 -820})

c) p_0 = 820-\frac{820}{e^6}

Step-by-step explanation:

For this case we have the following differential equation:

\frac{dp}{dt}=\frac{1}{2} (p-820)

And if we rewrite the expression we got:

\frac{dp}{p-820}= \frac{1}{2} dt

If we integrate both sides we have:

ln|P-820|= \frac{1}{2}t +c

Using exponential on both sides we got:

P= 820 + P_o e^{1/2t}

Part a

For this case we know that p(0) = 770 so we have this:

770 = 820 + P_o e^0

P_o = -50

So then our model would be given by:

P(t) = -50e^{1/2t} +820

And if we want to find at which time the population would be extinct we have:

0=-50 e^{1/2 t} +820

\frac{820}{50} = e^{1/2 t}

Using natural log on both sides we got:

ln(\frac{82}{5}) = \frac{1}{2}t

And solving for t we got:

t = 2 *ln(\frac{82}{5}) =5.595

Part b

For this case we know that p(0) = p0 so we have this:

p_0 = 820 + P_o e^0

P_o = p_0 -820

So then our model would be given by:

P(t) = (p_o -820)e^{1/2t} +820

And if we want to find at which time the population would be extinct we have:

0=(p_o -820)e^{1/2 t} +820

-\frac{820}{p_0 -820} = e^{1/2 t}

Using natural log on both sides we got:

ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t

And solving for t we got:

t = 2 *ln(-\frac{820}{p_0 -820})

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

12 = 2 *ln(\frac{820}{820-p_0})

6 = ln (\frac{820}{820-p_0})

Using exponentials we got:

e^6 = \frac{820}{820-p_0}

(820-p_0) e^6 = 820

820-p_0 = \frac{820}{e^6}

p_0 = 820-\frac{820}{e^6}

8 0
3 years ago
Solve this problem using the Trigonometric identities (secA+1)(SecA-1)= tan^2A
sammy [17]

Step-by-step explanation:

( secA + 1)( sec A - 1)

Using the expansion

( a + b)( a - b) = a² - b²

Expand the expression

We have

sec²A + secA - secA - 1

That's

sec² A - 1

From trigonometric identities

<h3>sec²A - 1 = tan ²A</h3>

So we have the final answer as

<h3>tan²A</h3>

As proven

Hope this helps you

7 0
3 years ago
Read 2 more answers
1. (x^3 - 9 x^2 + 26 x - 24) ÷ (x - 4)
kkurt [141]

Answer:

1. = x^2 -5x+6

Step-by-step explanation:

2. = 2x^2 +5x+3

3. =2x^2 +(x-1) +((x)/x^2 -x+6)

5 0
3 years ago
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