Round 399,411 to the nearest hundredth thousandth

In a bag of oranges, the ratio of the good ones to the bad ones is 5:4. If the number of bad ones in the bag is 36, how many ora

iVinArrow [24]

Add the bad and good together. You get 81

5 0

3 years ago

Find the quotient. (6x 2 - x - 40) ÷ (5 + 2x) 8 - 3x 3x + 8 3x - 8

Artyom0805 [142]

Answer:

The quotient is 3x-8

Step-by-step explanation:

(6x^2-x-40)÷ (5 + 2x)

The above equation can be written as

(6x^2-x-40)÷ (2x + 5)

The division is shown in the figure below.

The quotient is 3x-8

4 0

2 years ago

Help me please!!! 45 points to whoever helps!!

kap26 [50]

I don’t know 775 re2566

7 0

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

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

6 0

3 years ago

If the first of three consecutive integers is subtracted from 138, the result is the sum of the second and third. What are the i

otez555 [7]

Answer:

First Integer = n = 45

Second Integer = n+1 = 45 + 1 = 46

And Third Integer = n+ 2 = 45 +2 = 47

Step-by-step explanation:

Let First integer = n

Second Integer = n+1

Third Integer = n+2

According to the question given (If the first of three consecutive integers is subtracted from 138, the result is the sum of the second and third) the equation will be:

138 - n = (n+1) + (n+2)

Solving the equation:

138 - n = n+1+n+2

138 - n = 2n+3

138 - 3 =2n +n

135 = 3n

135/3 = n

=> n= 45

So, First Integer = n = 45

Second Integer = n+1 = 45 + 1 = 46

And Third Integer = n+ 2 = 45 +2 = 47

8 0

3 years ago