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alexandr1967 [171]

3 years ago

12

Travelling only through the crust, surface waves arrive after body waves. though they arrive later than body waves, surface wave

s are predominantly responsible for the damage and destruction caused by

2 answers:

leva [86]3 years ago

4 0

I believe the answer would be volcanic eruptions.

riadik2000 [5.3K]3 years ago

3 0

Answer:

volcanic eruptions

Step-by-step explanation:

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A line passes through the points (1.4) and (2, 2). What is the equation of this line?

emmasim [6.3K]

Answer:

y = - 2x + 6

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (1, 4 ) and (x₂, y₂ ) = (2, 2 )

m = $\frac{2-4}{2-1}$ = $\frac{-2}{1}$ = - 2 , then

y = - 2x + c ← is the partial equation

To find c substitute either of the 2 points into the partial equation

Using (2, 2 ) , then

2 = - 4 + c ⇒ c = 2 + 4 = 6

y = - 2x + 6 ← equation of line

3 0

2 years ago

Read 2 more answers

In a bag there are pink buttons, yellow buttons and blue buttons.

larisa86 [58]

Answer:

do you have any options for this question just to check??

8 0

2 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

3 years ago

Which graph represents the function of f(x) = 9x 2 -36/3x+6 Please help

UkoKoshka [18]

The function is:

         (9x^2 - 36)
f(x) = ---------------------
                3x + 6

You can simplify that function to a linear function for all x for which 3x + 6 ≠ 0

=> x ≠ - 2.

So, for x ≠ - 2, you can do:

         (9x^2 - 36)
f(x) = --------------------- =
               3x + 6

          9(x^2 - 4)
f(x) = ---------------------
                3(x + 2)

         3(x + 2)(x - 2)
f(x) = --------------------- = 3(x - 2) = 3x - 6
                 (x + 2)

So, the graph is a right line that intercepts the y-axis at - 6 and the x-axis at x = 2, excluding x = -2 as the function is not defined for x = -2. That is the second graph of the second picture.

6 0

3 years ago

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.

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

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

hence the complete solution

y = C.F. + P.I.

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

7 0

3 years ago