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lara [203]
2 years ago
13

Use: nCk=n!/k!(n-k)!

Mathematics
1 answer:
alekssr [168]2 years ago
3 0

Answer:

47

Step-by-step explanation:

7x2=14-10=4 so 47 Is an answer

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Find the length and perimeter of a rectangle if its width is 19 m and its area is 475 m?.
slava [35]
Answer = 25

Explanation

Area divided by width: 475 / 19 = 25

Therefore answer is: Width - 25m
8 0
2 years ago
A farmer was counting his cows and chickens. He counted 13 heads and 36 legs. How many cows does he have?
melomori [17]
<span>Cows (C)→ 4 legs
Chickens (c) → 2 legs
C+c=13    |×4
4C+2c=36

4C+4c=52
4C+2c=36
--------------- (-)
       2c=16
       c=8 chickens

C=13-8
C=5 cows
</span>
3 0
3 years ago
A student got an 80% on a test that had 40 questions on it. How many questions did the student get correct?
evablogger [386]
The student got 32 questions correct.
40 x 0.8 = 32
6 0
3 years ago
Find two values of c in (− π/ 4 , π /4) such that f(c) is equal to the average value of f(x) = 2 cos(2x) on ( − π/ 4 , π/ 4 ). R
pentagon [3]

Answer:

c₁ = 1/2 cos⁻¹ (2/π) = 0.44

c₂ = -1/2 cos⁻¹ (2/π) = -0.44

Step-by-step explanation:

the average value of f(x)=2 cos(2x) on ( − π/ 4 , π/ 4 ) is

av f(x) =∫[2*cos(2x)] dx /(∫dx) between limits of integration − π/ 4 and π/ 4

thus

av f(x) =∫[cos(2x)] dx /(∫dx) = [sin(2 * π/ 4 ) - sin(2 *(- π/ 4 )] /[ π/ 4 -  (-π/ 4)]

= 2*sin (π/2) /(π/2) = 4/π

then the average value of f(x) is 4/π . Thus the values of c such that f(c)= av f(x) are

 4/π = 2 cos(2c)  

2/π = cos(2c)

c = 1/2 cos⁻¹ (2/π) = 0.44

c= 0.44

since the cosine function is symmetrical  with respect to the y axis then also c= -0.44 satisfy the equation

thus

c₁ = 1/2 cos⁻¹ (2/π) = 0.44

c₂ = -1/2 cos⁻¹ (2/π) = -0.44

4 0
3 years ago
(15 pts) 4. Find the solution of the following initial value problem: y"-10y'+25y = 0 with y(0) = 3 and y'(0) = 13
jolli1 [7]

Answer:

y(x)=3e^{5x}-2xe^{5x}

Step-by-step explanation:

The given differential equation is y''-10y'+25y=0

The characteristics equation is given by

r^2-10r+25=0

Finding the values of r

r^2-5r-5r+25=0\\\\r(r-5)-5(r-5)=0\\\\(r-5)(r-5)=0\\\\r_{1,2}=5

We got a repeated roots. Hence, the solution of the differential equation is given by

y(x)=c_1e^{5x}+c_2xe^{5x}...(i)

On differentiating, we get

y'(x)=5c_1e^{5x}+5c_2xe^{5x}+c_2e^{5x}...(ii)

Apply the initial condition y (0)= 3 in equation (i)

3=c_1e^{0}+0\\\\c_1=3

Now, apply the initial condition y' (0)= 13 in equation (ii)

13=5(3)e^{0}+0+c_2e^{0}\\\\13=15+c_2\\\\c_2=-2

Therefore, the solution of the differential equation is

y(x)=3e^{5x}-2xe^{5x}

5 0
3 years ago
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