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3 years ago

Solve1 over 25 = 5^x ^+ 4. x = -7/2 x = −6 x = 9/2 x = 2

1 answer:

6 0

The equation seems to be

   1
-------- = 5 ^ (x + 4)
25

In that case, this is the solution, stept by step:

1) factor 25: 25 = 5^2

      1
=> ----------- = 5^(x + 4)
       5^2

2) invert 5^2

=> 5^(-2) = 5^(x+4)

3) Given that the bases are equal, the exponents also have to be equal:

=> - 2 = x + 4

3) transpose +4:

=> - 2 - 4 = x

=> x = - 6

Answer: - 6

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Simplify (sec θ – tan θ)2

rusak2 [61]

Answer:

2 sec (θ ) − 2 tan ( θ )

Step-by-step explanation:

4 0

3 years ago

If Bob The Baker can bake 12 pies in 8 hours, how many pies can he bake in a day (Hint: How many hours are in a day?

shusha [124]

He can make 36 pies in a day because there are 24 hours in a day and
24/8=3
3•12=36

6 0

3 years ago

Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,

posledela

Answer:

number of multiplies is n!
n=10, 3.6 ms
n=15, 21.8 min
n=20, 77.09 yr
n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

mpy[n] = n·mp[n-1]

mpy[2] = 2

So, ...

mpy[n] = n! . . . n ≥ 2

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0

3 years ago

Pls helpp

irina1246 [14]

you will need 2 busses to only transport the boys.
Mark is at (1 + 5/6) miles of his house.

<h3>How many buses would it take to carry only the boys?</h3>

We know that there are (3 + 1/2) groups, such that each group fill one bus.

2/5 of the students are boys, then the number of groups that we can make only with boys is:

(2/5)*(3 + 1/2) = 6/5 + 1/5 = 7/5 = 5/5 + 2/5 = 1 + 2/5

Then you can make one and a little less than a half of a group, which means that you need 1 and 2/5 of a buss to transport the boys, rounding that to the a whole number, you will need 2 busses to only transport the boys.

<h3>How far is Mark from his house?</h3>

The original distance is:

D = (2 + 3/4) miles.

But Mark only covers 2/3 of that distance, then we have:

d = (2/3)*D = (2/3)*(2 + 3/4) miles = (4/3 + 2/4) miles

d = (4/3 + 1/2) miles = (8/6 + 3/6) miles = (1 + 5/6) miles

Mark is at (1 + 5/6) miles of his house.

If you want to learn more about mixed numbers:

brainly.com/question/21610929

#SPJ1

7 0

2 years ago

Prove that at a party where some people shake hands, the number of people who shake hands with an odd number of people is an eve

Kobotan [32]

If there are n people, each person could shake hands with 0 people, 1 person, 2 people,... on up to shaking hands with n − 1
people. Count how many different answers there are to asking the person the question "How many hands did you shake?" How many people are there? If the people are the pigeons, and the possible answers to the question "how many hands did you shake" are the holes, can we conclude anything yet? No? How about now noticing that at least one of the holes "I shook hands with noone" or "I shook hands with everyone" has to be empty... now what?

"Since there are more pigeons than holes there must be a hole with at least two pigeons in the same hole" Now, replace the word "pigeons" and "holes" with the appropriate terms for the context of your specific question, remember we are talking about people and number of handshakes they participated in.

5 0

3 years ago