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Serggg [28]
2 years ago
9

Please help me answer this question genius

Mathematics
1 answer:
kolbaska11 [484]2 years ago
7 0

Answer:

  x ∈ {-0.465, 1.014}

Step-by-step explanation:

The equation can be cast in the form f(x) = 0, and solved easily using a graphing calculator. That shows x ≈ -0.465 and x ≈ 1.014. The same calculator can iterate the roots to full calculator precision.

__

The equation can be made a quadratic by the substitution ...

  z = e^(2x)

Then we have ...

  2\cosh(2x)-\sinh(2x)=4\\\\2\cdot\dfrac{e^{2x}+e^{-2x}}{2}-\dfrac{e^{2x}-e^{-2x}}{2}-4=0\qquad\text{use exponential identities}\\\\2z+2z^{-1}-z+z^{-1}-8=0\qquad\text{multiply by 2, substitute z}\\\\z^2 -8z+3=0\qquad\text{multiply by z}\\\\z=\dfrac{8\pm\sqrt{(-8)^2-4(1)(3)}}{2(1)}=4\pm\sqrt{13}\\\\x=\dfrac{\ln(z)}{2}=\dfrac{\ln(4\pm\sqrt{13})}{2}\approx\{-0465133,1.014439\}

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How would I solve this?
trasher [3.6K]

You had the right idea using the Pythagorean theorem to solve for b.

Problem is for that triangle to work, the 5 and the 2√2 would have to switch places. The length of a leg cannot be larger than the length of the hypotenuse for it to truly be a right triangle.

Pythagorean theorem only works for the right triangles. Only way to "solve this problem would be to bring in complex numbers.

5² + b² = (2√2)²

25 + b² = 2²(√2)²

25 + b² = 4(2)

25 + b² = 8

b² = 8 - 25

b² = - 17

b = √-17

b= (√17i)

Then the problem with THIS is a measurement/distance cannot be negative... which goes against exactly what that complex number i is.

8 0
3 years ago
shirley has $540 un her bank account. She withdraws $35 each week to cover her expenses. write an equation that relates the amou
Veseljchak [2.6K]
540-35x= y

540 is the amount she has minus 35 and x is the weeks she withdrawn the money and y is the answer.
6 0
3 years ago
Pleeeeeeeeeeeeeeeeeease help
IgorLugansk [536]

Answer:

(-5/7)^17

Step-by-step explanation:

When multiplying numbers to a certain power, you will add the powers together. When dividing numbers to a certain power, you will subtract the powers.

8 0
3 years ago
Problem 4: Solve the initial value problem
pishuonlain [190]

Separate the variables:

y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx

Separate the left side into partial fractions. We want coefficients a and b such that

\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}

\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}

\implies 1 = a(y-2)+b(y+1)

\implies 1 = (a+b)y - 2a+b

\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13

So we have

\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx

Integrating both sides yields

\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx

\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C

\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C

\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C

\dfrac{y-2}{y+1} = e^{3x + C}

\dfrac{y-2}{y+1} = Ce^{3x}

With the initial condition y(0) = 1, we find

\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12

so that the particular solution is

\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}

It's not too hard to solve explicitly for y; notice that

\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}

Then

1 - \dfrac3{y+1} = -\dfrac12 e^{3x}

\dfrac3{y+1} = 1 + \dfrac12 e^{3x}

\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}

y+1 = \dfrac6{2+e^{3x}}

y = \dfrac6{2+e^{3x}} - 1

\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}

7 0
2 years ago
4. 3x + 7 + 5x = 16 + 7 then x =​
VARVARA [1.3K]

4.3x+5x=16+7-(7)

9.3x=16

x= 1.72

give me a brainliest if possible. please.. tnx

8 0
3 years ago
Read 2 more answers
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