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3 years ago

What is the slope of the line through (0,-1) and (2,3)?

Mathematics

1 answer:

aniked [119]3 years ago

3 0

Slope = (y2-y1)/(x2-x1)
(3+1)/(2-0) = 4/2 = 2
The slope is 2

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Given that f(x) is even and g(x) is odd, determine whether each function is even, odd, or neither. (f • g)(x) = ? (g • g)(x) = ?

vekshin1

Answer:

Step-by-step explanation:

If f(x) is even, then f(-x) = f(x).

If g(x) is odd, then g(-x) = -g(x).

(f · g)(x) = f(x) · g(x)

Check:

(f · g)(-x) = f(-x) · g(-x) = f(x) · [-g(x)] = -[f(x) · g(x)] = -(f · g)(x)

(f · g)(-x) = -(f · g)(x) - odd

(g · g)(x) = g(x) · g(x)

Check:

(g · g)(-x) = g(-x) · g(-x) = [-g(x)] · [-g(x)] = g(x) · g(x) = (g · g)(x)

(g · g)(-x) = (g · g)(x) - even

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nadya68 [22]

You have to take the percentage of the original price and subtract it to find the reduced price

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Find the Laplace transformation of each of the following functions. In each case, specify the values of s for which the integral

MAVERICK [17]

Answer:

a. $\frac {2} {s-1}$ converges to s> 1.

b. $\frac{3}{e^3 \left(s-5 \right)}$ converges to s> 5.

c. $- \frac {2}{s + 3}$ converges to s> - 3.

d. $\frac {s}{s^2 + 25}$ converges to s> 0.

e. $\frac {10} {s^2 + 1}$ converges even s> 0.

f. $\frac {12}{s^2 + 4}$ converges to s> 0.

g. $-\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4}$ converges to s> 0.

h. $\frac {1} {s ^ 2 + 4}$ converges to s> 0.

Step-by-step explanation:

a. $L \left\{2e^t \right\} = 2L \left\{e^t \right\} = 2 \cdot \frac {1} {s-1} = \frac {2} {s-1}$ converges to s> 1.

b. $L \left\{3e^{5t-3} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = \frac{3}{e^3 \left(s-5 \right)}$ converges to s> 5.

c. $L \left\{-2e^{-3t} \right\} = -2L \left\{e^{-3t} \right\} = - \frac {2}{s + 3}$ converges to s> - 3.

d. $L \left\{\cos\left (5t \right)\right\} = \frac {s}{s^2 + 25}$ converges to s> 0.

e. $L \left\{10 \sin\left(t\right)\right\} = 10L\left\{\sin\left(t\right)\right\} = \frac {10} {s^2 + 1}$ converges even s> 0.

f. $L \left\{6\sin \left(2t \right) \right\} = 6L\left\{\sin\left (2t\right)\right\} = \frac {12}{s^2 + 4}$ converges to s> 0.

g. $L \left\{-5\cos\left(2t + 1\right) \right\} = -5L\left\{\cos\left(2t + 1 \right)\right\} = -\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4}$ converges to s> 0.

h. $L\left\{\sin \left(t\right)\cos \left(t\right)\right\} = L\left\{\sin\left(2t\right)\frac{1}{2}\right\} =\frac{1}{2}\cdot \frac{2}{s^2+4} = \frac {1} {s ^ 2 + 4}$ converges to s> 0.

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Answer:

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