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4 years ago

I need help solving this equation -3=9/(5-2k)/5

Mathematics

1 answer:

Amanda [17]4 years ago

3 0

Solve for k by simplifying both sides of the equation, then isolating the variable.

k= 14/5

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The number line shows the locations of points P 0. R. and S 털 HR & -7 -6 = Which points have a distance of 5 units between t

Svetradugi [14.3K]

Answer: point P and point S

Step-by-step explanation:

took the test

6 0

2 years ago

By copying and completing the working below, work out 3 + 1 16 Give your answer as a fraction in its simplest form.

Ksenya-84 [330]

Answer:

Assuming the question is: 3 + 1 16 [I see no "working below."

Step-by-step explanation:

3 + 1 16

48/16 + (16/16 + 1/16) [Make all numbers into fractions using 16 as the denominator. E.g. 48/16 = 3.

Add: 48/16 + (16/16 + 1/16)

(48+16+1)/16

=63/16

[Also equal to 3 15/16]

7 0

3 years ago

Directions: Determine how much Marco would be paid

lidiya [134]

Answer:

766.87 is how much he is paid in all.

Step-by-step explanation:

2.5% of 75 is 1.875. 1.875 times 9 is 16.875. 75 times ten is 750. add together to get the whole.

7 0

3 years ago

Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.

igor_vitrenko [27]

Answer:

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

$m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7$

So, the solution of the equation is

$y = Ae^{m_{1} t} + Be^{m_{2} t}$

where m₁ = 3 and m₂ = -7.

So,

$y = Ae^{3t} + Be^{-7t}$

Also,

$y' = 3Ae^{3t} - 7e^{-7t}$

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

$y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1 (1)$

$y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}$

Substituting A into (1) above, we have

$\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1 \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}$

Substituting B into A, we have

$A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}$

Substituting A and B into y, we have

$y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

So the solution to the differential equation is

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

6 0

4 years ago

Solve for x: x /6−y/3= 1<br> SHOW WORK

Dmitriy789 [7]

$\dfrac x6 - \dfrac y3 =1\\\\\\\implies \dfrac x6 - \dfrac{2y}6 = 1\\\\\\\implies \dfrac{x-2y}6 =1\\\\\\\implies x -2y =6\\\\\\\implies x =6+2y\\\\\\\implies x = 2(3 +y)$

3 0

2 years ago