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7nadin3 [17]
3 years ago
7

So much helpfucjfjfjfufufufhfbfbfhfbfbf fufjfbf ?

Mathematics
1 answer:
lara [203]3 years ago
6 0

Answer:

What u should do in this exercice?

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xYou volunteer to help drive children at a charity event to the​ zoo, but you can fit only 6 of the 16 children present in your
brilliants [131]

There are 8008 groups in total, in other to drive the children

<h3>How to determine the number of groups?</h3>

From the question, we have

  • Total number of children, n = 16
  • Numbers to children at once, r = 6

The number of group of children that could be carried at once is calculated using the following combination formula

Total = ⁿCᵣ

Where

n = 16 and r = 6

Substitute the known values in the above equation

Total = ¹⁶C₆

Apply the combination formula

ⁿCᵣ = n!/(n - r)!r!

So, we have

Total = 16!/10!6!

Evaluate

Total = 8008

Hence, the number of groups is 8008

Read more about combination at

brainly.com/question/11732255

#SPJ1

8 0
1 year ago
A line has a y-intercept of -2, but has no x-intercept. Describe
frutty [35]
If it has no x-intercept, it is a horizontal line going through 0,-2. hope this helps
6 0
3 years ago
Read 2 more answers
PLEASEE HELPPP
Andre45 [30]
It would be A) The angles are angle bisectors
I am 90% sure
5 0
3 years ago
Read 2 more answers
Gwen has recently purchased a cottage. She has arranged
Aleksandr-060686 [28]
A) 0.8
B) 0.24
C) 0.56
D) 0.44

For A, since 1 in 5 is dry, 4 in 5 are not; 4/5 = 0.8

For B, we know that 30%, 0.3, of the ones that are not dry are contaminated:
0.3(0.8) = 0.24

For C, 100%-30%=70% of wells that are not dry are not contaminated; 0.7(0.8) = 0.56

For D, she will either have contaminated water or no water:
0.24+0.2 = 0.44
5 0
3 years ago
How do i solve that question?
yawa3891 [41]

a) The solution of this <em>ordinary</em> differential equation is y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2   } }.

b) The integrating factor for the <em>ordinary</em> differential equation is -\frac{1}{x}.

The <em>particular</em> solution of the <em>ordinary</em> differential equation is y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}.

<h3>How to solve ordinary differential equations</h3>

a) In this case we need to separate each variable (y, t) in each side of the identity:

6\cdot \frac{dy}{dt} = y^{4}\cdot \sin^{4} t (1)

6\int {\frac{dy}{y^{4}} } = \int {\sin^{4}t} \, dt + C

Where C is the integration constant.

By table of integrals we find the solution for each integral:

-\frac{2}{y^{3}} = \frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32} + C

If we know that x = 0 and y = 1<em>, </em>then the integration constant is C = -2.

The solution of this <em>ordinary</em> differential equation is y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2   } }. \blacksquare

b) In this case we need to solve a first order ordinary differential equation of the following form:

\frac{dy}{dx} + p(x) \cdot y = q(x) (2)

Where:

  • p(x) - Integrating factor
  • q(x) - Particular function

Hence, the ordinary differential equation is equivalent to this form:

\frac{dy}{dx} -\frac{1}{x}\cdot y = x^{2}+\frac{1}{x} (3)

The integrating factor for the <em>ordinary</em> differential equation is -\frac{1}{x}. \blacksquare

The solution for (2) is presented below:

y = e^{-\int {p(x)} \, dx }\cdot \int {e^{\int {p(x)} \, dx }}\cdot q(x) \, dx + C (4)

Where C is the integration constant.

If we know that p(x) = -\frac{1}{x} and q(x) = x^{2} + \frac{1}{x}, then the solution of the ordinary differential equation is:

y = x \int {x^{-1}\cdot \left(x^{2}+\frac{1}{x} \right)} \, dx + C

y = x\int {x} \, dx + x\int\, dx + C

y = \frac{x^{3}}{2}+x^{2}+C

If we know that x = 1 and y = -1, then the particular solution is:

y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}

The <em>particular</em> solution of the <em>ordinary</em> differential equation is y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}. \blacksquare

To learn more on ordinary differential equations, we kindly invite to check this verified question: brainly.com/question/25731911

3 0
2 years ago
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