Solve for b. ab+c=d B=a+c/d B=a/(c-d) B=(d-c)/a

Mathematics

1 answer:

vekshin13 years ago

3 0

Answer:

b = (d-c)/a

Step-by-step explanation:

ab+c=d

Subtract c from each side

ab+c-c=d-c

ab = d-c

Divide by a

ab/a = (d-c)/a

b = (d-c)/a

You might be interested in

What is (-2)(-5)? Dhdjfhfhhdhd

lbvjy [14]

Answer:

$( - 2)( - 5) \\ - 2 \times - 5 \\ = - 10$

<h2>please mark me brainliest</h2>

6 0

3 years ago

A line contains points M(1, 3) and N(5, 0). What is the slope of MN? -4/3 -3/4 3/4 4/3

german

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Solve the following ODE's: c) y* - 9y' + 18y = t^2

Nastasia [14]

Answer:

y = $C_1e^{3t}+C_2e^{6t}$ + $\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

Step-by-step explanation:

y''- 9 y' + 18 y = t²

solution of ordinary differential equation

using characteristics equation

m² - 9 m + 18 = 0

m² - 3 m - 6 m+ 18 = 0

(m-3)(m-6) = 0

m = 3,6

C.F. = $C_1e^{3t}+C_2e^{6t}$

now calculating P.I.

$P.I. = \frac{t^2}{D^2 - 9D +18}$

$P.I. = \dfrac{t^2}{(D-3)(D-6)}\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1-\frac{D}{6})^{-1}(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1+\frac{D}{6}+\frac{D^2}{36}+....)(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(1+\frac{D}{3}+\frac{D^2}{9}+....)(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$