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3 years ago

The sum of two numbers is 54. their difference is 34. find the numbers.

1 answer:

4 0

X + y = 54
x - y = 34
----------------add
2x = 88
x = 88/2
x = 44

x + y = 54
44 + y = 54
y = 54 - 44
y = 10

so ur 2 numbers are 44 and 10

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postnew [5]

Answer:

the first one I guess

Step-by-step explanation:

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2 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

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

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3 years ago

True or False: A square is a specific type of parallelogram?

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False, they are both quadrilaterals though.

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2 years ago

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Step-by-step explanation:

By going on a calculator and typing 81 then click divide then press 13.5

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3 years ago

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For the month of March in a certain city, 57% of the days are cloudy. Also in the month of March in the same city, 55% of the da

oksian1 [2.3K]

Answer: 0.9649

Step-by-step explanation:

Let A denote the event that the days are cloudy and B denotes the event that the days are rainy.

Given : For the month of March in a certain city, the probability that days are cloudy : $P(A)=0.57$

Also in the month of March in the same city,, the probability that the days are cloudy and rainy : $P(A\cap B)=0.55$

Now by using the conditional probability, the probability that a randomly selected day in March will be rainy if it is cloudy will be :-

$P(B|A)=\dfrac{P(A\cap B)}{P(A)}$

$\Rightarrow\ P(B|A)=\dfrac{0.55}{0.57}\\\\=0.964912280702\approx0.9649\ \ \text{[Rounded to four decimal places.]}$

Hence, the probability that a randomly selected day in March will be rainy if it is cloudy = 0.9649

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3 years ago