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Molodets [167]
3 years ago
10

Leeza is 6 years older than two times her brother Sammy’s age. Let s represent Sammy’s age.

Mathematics
2 answers:
m_a_m_a [10]3 years ago
5 0

✧・゚: *✧・゚:*hello there!*:・゚✧*:・゚✧

✧here is your answer:

.・゜゜・d. 2s + 6・゜゜・.

♥⛧i hope this helps!!

✧༺♥༻∞lmk if you need help on anything else!!∞༺♥༻✧

❀-kay♬

Anna007 [38]3 years ago
4 0
D.2s + 6.
Hope you have a nice dayyyyyyyyyy
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3 years ago
(a^2-ax+bx-ab)/(a^2+ax-bx-ab)*(a^2+ax+ab+bx)/(a^2-ax-ab+bx)
Vanyuwa [196]
\frac{a^2-ax+bx-ab}{a^2+ax-bx-ab} \cdot \frac{a^2+ax+ab+bx}{a^2-ax-ab+bx} = \frac{1}{a(a+x)-b(x+a)} \cdot (a(a+x)+b(a+x)) = \frac{1}{a-b} \cdot (a+b) = \frac{a+b}{a-b}.
8 0
3 years ago
Any help, please :)
Nataly_w [17]

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y=x+6

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3 0
3 years ago
Read 2 more answers
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aliina [53]
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4 0
4 years ago
Find the solution of the differential equation that satisfies the given initial condition. y' tan x = 3a + y, y(π/3) = 3a, 0 &lt
Paladinen [302]

Answer:

y(x)=4a\sqrt{3}* sin(x)-3a

Step-by-step explanation:

We have a separable equation, first let's rewrite the equation as:

\frac{dy(x)}{dx} =\frac{3a+y}{tan(x)}

But:

\frac{1}{tan(x)} =cot(x)

So:

\frac{dy(x)}{dx} =cot(x)*(3a+y)

Multiplying both sides by dx and dividing both sides by 3a+y:

\frac{dy}{3a+y} =cot(x)dx

Integrating both sides:

\int\ \frac{dy}{3a+y} =\int\cot(x) \, dx

Evaluating the integrals:

log(3a+y)=log(sin(x))+C_1

Where C1 is an arbitrary constant.

Solving for y:

y(x)=-3a+e^{C_1} sin(x)

e^{C_1} =constant

So:

y(x)=C_1*sin(x)-3a

Finally, let's evaluate the initial condition in order to find C1:

y(\frac{\pi}{3} )=3a=C_1*sin(\frac{\pi}{3})-3a\\ 3a=C_1*\frac{\sqrt{3} }{2} -3a

Solving for C1:

C_1=4a\sqrt{3}

Therefore:

y(x)=4a\sqrt{3}* sin(x)-3a

3 0
4 years ago
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