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lilavasa [31]
4 years ago
15

I need to know how to go by completing this

Mathematics
1 answer:
Pepsi [2]4 years ago
7 0

Answer:it would be ax+by=c in this case add x and x squared and then do 6x squared then add 18 and your answer would be C

Step-by-step explanation:

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Which statement is true? Question 17 options: A) |–3| = 3 and |–4| = –4 B) –|3| = 3 and |4| = 4 C) |–3| = 3 and –|–4| = –4 D) |–
Mkey [24]
The second one because 4d x c is the correct answer
8 0
3 years ago
10,996 rounded to the neart thousand
Sholpan [36]
<h3>Rounding it to the nearest thousand:</h3>

11,000

<h2>Answer:</h2>

11,000

3 0
4 years ago
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Need help with this thank you 10 points
pochemuha
250 you have to divide, btw im hella wrong

8 0
4 years ago
Find the solution of the initial value problem y Superscript prime prime Baseline plus 4 y Superscript prime Baseline plus 5 y e
sergiy2304 [10]

Answer:

y(t) =  - 5 {e}^{2\pi - 2t} \cos(t)

Step-by-step explanation:

The given initial value problem is

y''+4y'+5=0

y( \frac{ \pi}{2} ) = 0

y'( \frac{ \pi}{2} ) = 5

The corresponding characteristic equation is

{m}^{2}  + 4m + 5 = 0

m =  - 2 \pm \: i

The general solution becomes:

y(t) = A {e}^{ - 2t}  \cos(t)  +  B {e}^{ - 2t}  \sin(t)

We differentiate to get:

y'(t) =  - A {e}^{ - 2t}  \sin(t)  - 2 A {e}^{ - 2t}  \cos(t)  + B {e}^{ - 2t} \cos(t)   - 2B {e}^{ - 2t}  \sin(t)

We apply the initial conditions to get;

y( \frac{\pi}{2} ) = A {e}^{ - 2\pi}  \cos( \frac{\pi}{2} )  +  B {e}^{ - 2\pi}  \sin( \frac{\pi}{2} )

A {e}^{ - 2\pi} (0 )  +  B {e}^{ - 2\pi}  ( 1)  = 0

B  = 0

Also;

y'( \frac{\pi}{2} ) =  - A {e}^{ - 2\pi}  \sin( \frac{\pi}{2} )  - 2 A {e}^{ - 2\pi}  \cos( \frac{\pi}{2} )  + B {e}^{ - 2\pi} \cos( \frac{\pi}{2} )   - 2B {e}^{ - 2\pi}  \sin( \frac{\pi}{2} )

- A {e}^{ - 2\pi} ( 1 )  - 2 A {e}^{ - 2\pi} ( 0 )  + B {e}^{ - 2\pi}( 0)   - 2B {e}^{ - 2\pi} ( 1 )  = 5

- A {e}^{ - 2\pi}     - 2B {e}^{ - 2\pi} = 5

But B=0

- A {e}^{ - 2\pi}  = 5

A = 5{e}^{2\pi}

Therefore the particular solution is

y(t) =  - 5 {e}^{2\pi} {e}^{ - 2t}  \cos(t)  +  0 \times {e}^{ - 2t}  \sin(t)

y(t) =  - 5 {e}^{2\pi - 2t} \cos(t)

4 0
4 years ago
PLEASE HELP!!!!
Tasya [4]

Answer:

y = g(x) = 1 +  x

Now, the range of a function is the set of the possible values of y.

In this function, a linear function, y can be any real number, so the range of this function is { y ∈ ℝ )

y = h(x) = x^2 + 2*x

This is a quadratic function, as the leading coefficient is positive, we know that the arms of the function go up.

For a quadratic function y = a*x^2 + b*x + c

The minimum is at:

x = -b/2a.

In this case, b = 2 and 1 = 1

the minimum is at:

x = -2/2 = -1

The minimum is:

y = h(-1) = 1^2 +2*(-1) = -1

Then the range of this function is

{y ∈ ℝ ≥ -1 )

The composite functions are:

g∘h = g(h(x)) = 1 + x^2 + 2*x

The minimum is still at x = -1

g∘h(-1) = 1 + -1^2 + 2*-1 = 0

Then the range of this function is:

{y ∈ ℝ ≥ 0 )

The other composition is:

h∘g = h(g(x)) =  (1 + x)^2 + 2*(1 + x) = 1 + 2*x + x^2 + 2 + 2*x

h∘g = x^2 + 4*x + 3

Here the minimum is at:

x = -4/2*1 = -2

h∘g(-2) = (-2)^2 + 4*-2 + 3 = 4 - 8 + 3 = -1

The range is:

{y ∈ ℝ ≥ -1 )

5 0
3 years ago
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