Which is the measure of line segment AB? A)10 feet B)15 feet C)20 feet D)100 feet

Mathematics

2 answers:

nalin [4]2 years ago

6 0

Answer:

AB= 20 feet

Step-by-step explanation:

Given triangle ABC is a right angle triangle

Use trigonometric ratio to find the length of the Side AB

AB is the hypotenuse

Sin = opposite side divide by hypotenuse

$sin(B)= \frac{AC}{AB}$

$sin(30)= \frac{10}{AB}$

Cross multiply it

$sin(30) \cdot AB= 10$

$AB= \frac{10}{sin(30)}$

AB= 20 feet

zavuch27 [327]2 years ago

5 0

Find the last angle:
180-(30+90)
180-(120)
60

Evaluate using sine rule:
$\frac{c}{sinC}$ = $\frac{b}{sinB}$

c= $\frac{bsinC}{sinB}$

= $\frac{10sin90°}{sin30°}$

=20

The length of line segment AB is 20 feet (answer choice C)

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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.

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

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