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Monica [59]
2 years ago
7

Which set of instructions below best describes the steps to solve the following problem?

Mathematics
2 answers:
neonofarm [45]2 years ago
6 0
The answer is B I did the math on a calculator
user100 [1]2 years ago
3 0

Answer:

the answer is B

Step-by-step explanation:

because it is

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Question 3 Unsaved
bazaltina [42]

Answer:

Step-by-step explanation:

DE given is

y''+8y'+15y=0, y(0)=0, y'(0)=1

Take Laplace on the DE

We get

s^2 Y(s) -sY(0) -y'(0)+8(sY(s)-y(0))+15Y(s) =0\\s^2 Y(s) -s(0) -1+8(sY(s)-0)+15Y(s) =0\\Y(s)(s^2+8s+15)-1=0\\Y(s) = \frac{1}{s^2+8s+15}

Simplify to get

Y(s) = \frac{1}{2}[ {\frac{1}{s+3} -\frac{1}{s+5} }]

Take inverse

y(t) = \frac{e^{-3t} -e^{-5t}}{2}

8 0
3 years ago
Identify the domain and range
UkoKoshka [18]

Answer:

Domain: 1 ≤ x ≤ 4

Range : 1 ≤ f(x) ≤ 4

Step-by-step explanation:

The domain of a function f(x) is the limit within which the values of x varies.

Here, in the graph, it shows that the maximum value of x is 4 and the minimum value of x is 1.

Therefore, the domain of the function is 1 ≤ x ≤ 4

Again the range of a function f(x) is the limit within which the values of f(x) vary.

Here, the graph shows that the maximum value of f(x) is 4 and the minimum value of f(x) is 1.

Therefore, the range of the function f(x) is 1 ≤ f(x) ≤ 4. (Answer)

6 0
3 years ago
Let f(x) = (x − 3)−2. Find all values of c in (1, 7) such that f(7) − f(1) = f '(c)(7 − 1). (Enter your answers as a comma-separ
Sidana [21]

Answer:

This contradicts the Mean Value Theorem since there exists a c on (1, 7) such that f '(c) = f(7) − f(1) (7 − 1) , but f is not continuous at x = 3

Step-by-step explanation:

The given function is

f(x)=(x-3)^{-2}

When we differentiate this function with respect to x, we get;

f'(x)=-\frac{2}{(x-3)^3}

We want to find all values of c in (1,7) such that f(7) − f(1) = f '(c)(7 − 1)

This implies that;

0.06-0.25=-\frac{2}{(c-3)^3} (6)

-0.19=-\frac{12}{(c-3)^3}

(c-3)^3=\frac{-12}{-0.19}

(c-3)^3=63.15789

c-3=\sqrt[3]{63.15789}

c=3+\sqrt[3]{63.15789}

c=6.98

If this function satisfies the Mean Value Theorem, then f must be continuous on  [1,7] and differentiable on (1,7).

But f is not continuous at x=3, hence this hypothesis of the Mean Value Theorem is contradicted.

 

3 0
3 years ago
Please help with my geometry
faust18 [17]

Answer:

www.mahtpapa.com

Step-by-step explanation:

5 0
3 years ago
Whats 9x9-18+18-20+2?
zhuklara [117]

Answer:

9x^9-18

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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