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3 years ago

How old will steve be in 2 years if k years ago he was w years old

Mathematics

2 answers:

Annette [7]3 years ago

5 0

(k+w)+2 would be the answer if it was an algebraic expression.

Tanya [424]3 years ago

4 0

He would be 15 year old old enough to school

You might be interested in

Help me what is X/5-3/7

yaroslaw [1]

Answer:

15/7 or 2.142

Step-by-step explanation:

x=3/7×5

x=15/7

3 0

2 years ago

F(x)=<img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2-4%7D" id="TexFormula1" title="\sqrt{x^2-4}" alt="\sqrt{x^2-4}" align="abs

hammer [34]

Solve for x when √(x ² - 4) = 1 :

√(x ² - 4) = 1

x ² - 4 = 1

x ² = 5

x = ±√5

We're looking at x ≤ 0, so we take the negative square root, x = -√5.

This means f (-√5) = 1, or in terms of the inverse of f, we have f ⁻¹(1) = -√5.

Now apply the inverse function theorem:

If f(a) = b, then (f ⁻¹)'(b) = 1 / f '(a).

We have

f(x) = √(x ² - 4) → f '(x) = x / √(x ² - 4)

So if a = -√5 and b = 1, we get

(f ⁻¹)'(1) = 1 / f ' (-√5)

(f ⁻¹)'(1) = √((-√5)² - 4) / (-√5) = -1/√5

The sign must be negative; see the attached plot, and take note of the negatively-sloped tangent line to the inverse of f at x = 1.

3 0

3 years ago

Please help me , thank you

Mama L [17]

Answer:

Step-by-step explanation:

The range are the values that y can take.

From the graph we can see that all defined values of y are below the x- axis

Thus range is y < 0 → b

5 0

3 years ago

Hello there!

djyliett [7]

2. f(x) = x - 2x² - 5 + 10x

=-2x² + 8x - 5

f'(x)= -4x + 8

4. y = 100(45x - 30 - 3x³ + 2x²)

= 100(-3x³ + 2x² + 45x -30)

= -300x³ + 200x² + 4500x - 3000

y' = -900x² + 400x + 4500

5 0

3 years ago

Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.

$y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7$

Substitute these into the ODE to see everything checks out:

$2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14$

5 0

3 years ago