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Dmitry [639]
3 years ago
14

How to do this question

Mathematics
2 answers:
Hoochie [10]3 years ago
5 0
312.5 x 13. Which is <span>4062.5. Then add 1562.5 to get 5625. So by 1998 there would be 5625 bald eagle pairs</span>
Dafna11 [192]3 years ago
3 0
You cannot solve for a unique solution to x and y 
You might be interested in
Thx in advance for the help
lubasha [3.4K]

Answer: x\geq 13

Step-by-step explanation:

Given

The inequality is  2x-7\geq 19

adding both side 7

2x-7+7\geq 19+7\\2x\geq 26

Multiply both sides by \frac{1}{2}

\dfrac{1}{2}\times 2x\geq 26\times \dfrac{1}{2}\\\\x\geq 13

the shaded region in the figure indicates the solution set.

6 0
3 years ago
A vendor has 22 balloons for sale: 11 are yellow, 4 are red, and 7 are green. A balloon is selected at random and sold. If the b
pochemuha

Answer:

10/21

Step-by-step explanation:

After the first balloon is gone, there is only 21 balloons left with 10 being yellow. Therefore, the is a 10/21 probability that it is yellow.

4 0
4 years ago
Read 2 more answers
In a lab experiment, 610 bacteria are placed in a petri dish. The conditions are such that the number of bacteria is able to dou
Alex777 [14]

Answer:

It would take 19 hours and 36 minutes until there are 1040 bacteria present.

Step-by-step explanation:

Given that in a lab experiment, 610 bacteria are placed in a petri dish, and the conditions are such that the number of bacteria is able to double every 23 hours, to determine how long would it be, to the nearest tenth of an hour, until there are 1040 bacteria present, the following calculation must be performed:

610X = 1040

X = 1040/610

X = 1.7049

2 = 23

1.7049 = X

1.7049 x 23/2 = X

39.2131 / 2 = X

19.6 = X

100 = 60

60 = X

60 x 60/100 = X

36 = X

Therefore, it would take 19 hours and 36 minutes until there are 1040 bacteria present.

5 0
3 years ago
Calculus 2 master needed; stuck on evaluating the integral please show steps <img src="https://tex.z-dn.net/?f=%5Cint%20%7Bsec%2
kakasveta [241]

Answer:

\int \sec(\frac{x}{2})\tan^5({\frac{x}{2}})dx=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)+C

Step-by-step explanation:

So we have the integral:

\int \sec(\frac{x}{2})\tan^5({\frac{x}{2}})dx

First, let's use substitution to get rid of the x/2. I'm going to use the variable y. So, let y be x/2. Thus:

y=\frac{x}{2}\\dy=\frac{1}{2}dx\\2dy=dx

Therefore, the integral is:

=2\int \sec(y)\tan^5(y)dy

Now, as you had done, let's expand the tangent term. However, let's do it to the fourth. Thus:

=2\int \sec(y)\tan^4(y)\tan(y)dy

Now, we can use a variation of the trigonometric identity:

\tan^2(y)+1=\sec^2(y)

So:

\tan^2(y)=\sec^2(y)-1

Substitute this into the integral. Note that tan^4(x) is the same as (tan^2(x))^2. Thus:

=2\int \sec(y)(\tan^2(y))^2\tan(y)dy\\=2\int \sec(y)(\sec^2(y)-1)^2\tan(y)dy

Now, we can use substitution. We will use it for sec(x). Recall what the derivative of secant is. Thus:

u=\sec(y)\\du=\sec(y)\tan(y)dy

Substitute:

2\int (\sec^2(y)-1)^2(\sec(y)\tan(y))dy\\=2\int(u^2-1)^2 du

Expand the binomial:

=2\int u^4-2u^2+1du

Spilt the integral:

=2(\int u^4 du+\int-2u^2du+\int +1du)

Factor out the constant multiple:

=2(\int u^4du-2\int u^2du+\int(1)du)

Reverse Power Rule:

=2(\frac{u^{4+1}}{4+1}-2(\frac{u^{2+1}}{2+1})+(\frac{u^{0+1}}{0+1}}))

Simplify:

=2(\frac{u^5}{5}-\frac{2u^3}{3}+u)

Distribute the 2:

=\frac{2u^5}{5}-\frac{4u^3}{3}+2u

Substitute back secant for u:

=\frac{2\sec^5(y)}{5}-\frac{4\sec^3(y)}{3}+2\sec(y)

And substitute back 1/2x for y. Therefore:

=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)

And, finally, C:

=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)+C

And we're done!

7 0
3 years ago
Read 2 more answers
julius and olivia collect coins. julius has 338 coins in his collection. the number of coins olivia has is 57 less than double t
anygoal [31]
1) Julius and Olivia collect coins. {Qualitative statement}

2) Julius (J) has 338 coins in his collection. {Quantitative statement}

J = 338

3) The number of coins Olivia (V) has is 57 less than double the number of coins Julius has. {Quantitative statement}

V = 2*J - 57

V = 2*338 - 57

V = 619

------

Olivia has 619 coins.
7 0
3 years ago
Read 2 more answers
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