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2 years ago

PLS HELP ILL MARK BRAINLEST PLUS ADDITIONAL POINTSS!!

Mathematics

2 answers:

Serga [27]2 years ago

7 0

Answer:

114

Step-by-step explanation:

15 x 2 + 9 x 2 + 11 x 3 + 11 x 3 = 114

(the bottom portion is equal to 3 x 11)

SpyIntel [72]2 years ago

3 0

11 plus 9 plus 11 is 31.

31 plus 15 plus 15 is 61.

61 plus 11 plus 11 is 83.

83 plus 9 plus 9 is 101.

13 minus 9 is 4.

So the answer is C 114 mm.

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The given two-parameter family is a solution of the indicated differential equation on the interval

mr_godi [17]

Answer:

I do 1 option for you as an example, you need to check the leftover by yourself.

Step-by-step explanation:

for d) y(0) = 0 and y'(pi) =0

$y(0) = C_1e^0cos(0)+ C_2 e^0 sin(0) = 0 \longrightarrow C_1 = 0$

$y(x) ' = C_1e^x cos(x) - C_1e^x sin (x) + C_2e^x sin(x) + C_2e^x cos(x)$

$y(\pi)'=C_1e^\pi cos(\pi)- C_1e^\pi sin(\pi)+ C_2e^\pi sin(\pi) + C_2e^\pi cos (\pi)$

Replace $C_1 = 0$ we have

$y'(\pi) = -C_2e^\pi = 0$

if and only if $C_2 =0$

Hence the given solution does not work.

then, d is NOT the correct answer.

3 0

2 years ago

To which subsets does 15/5 belong?

Alexus [3.1K]

Answer:

Whole, natural, integer, rational, real

Step-by-step explanation:

15/5 can be rewritten as 3.

7 0

2 years ago

Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be

Travka [436]

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C' (C' = kC)

taking exponents of both sides, we have

$L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt} (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k} - \frac{C"}{k} e^{kt}$

When t = 0, A(0) = 0 (since the forest floor is initially clear)

$A = \frac{L}{k} - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k} - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k} - \frac{C"}{k} e^{0}\\\frac{L}{k} = \frac{C"}{k} \\C" = L$

$A = \frac{L}{k} - \frac{L}{k} e^{kt}$

So, D = R - A =

$D = \frac{L}{k} - \frac{L}{k} - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}$

when t = 0(at initial time), the initial value of D =

$D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}$

4 0

2 years ago

How to turn this equation: <br> y= -1.02(x-3)^2+9.25<br> into factored form

MatroZZZ [7]

Answer:

$y=(102x-306-5\sqrt{3774})(102x-306+5\sqrt{3774})$

Step-by-step explanation:

You first equate it to zero to get:

$-1.02(x-3)^2+9.25=0$

Then solve using square root method

$-1.02(x-3)^2=-9.25$

$(x-3)^2=\frac{-9.25}{-1.02}$

$(x-3)^2=\frac{925}{102}$

$x=3\pm\sqrt{\frac{925}{102}}$

$x=3\pm{\frac{5\sqrt{3774}}{102}}$

$x={\frac{306\pm5\sqrt{3774}}{102}}$

Now work it backwards

$102x-(306\pm5\sqrt{3774})=0$

$(102x-306-5\sqrt{3774})(102x-306+5\sqrt{3774})=0$

Hence the factored form is:

$y=(102x-306-5\sqrt{3774})(102x-306+5\sqrt{3774})$

8 0

2 years ago

What is the equation of the line that passes through (4, 3) and (2, 2)?

anastassius [24]

(4,3)(2,2)
slope = (2 - 3) / (2 - 4) = -1/-2 = 1/2

y = mx + b
slope(m) = 1/2
use either of ur sets of points...(4,3)...x = 4 and y = 3
now we sub and find b, the y int
3 = 1/2(4) + b
3 = 2 + b
3 - 2 = b
1 = b

so ur equation is : y = 1/2x + 1 <==

4 0

3 years ago