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mars1129 [50]
3 years ago
7

A plain travels 3400 miles in 8 hours. How far will it travel in 6 hours at this rate

Mathematics
1 answer:
pashok25 [27]3 years ago
4 0

Answer:

2550 miles in 6 hours

Step-by-step explanation:

if you divide the miles and hours by each other you will get the m/ph (miles per hour)

3400 ÷ 8 = 425

now, that we know the plane travels 435 m/ph, then if we multiply it by 6 we will get the answer

425 × 6 = 2550

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Find the solution of the differential equation that satisfies the given initial condition. dy/dx=x/y , y(0)=-1
Charra [1.4K]

Separating variables, we have

\dfrac{dy}{dx} = \dfrac xy \implies y\,dy = x\,dx

Integrate both sides.

\displaystyle \int y\,dy = \int x\,dx

\dfrac12 y^2 = \dfrac12 x^2 + C

Given that y(0)=-1, we find

\dfrac12 (-1)^2 = \dfrac12 0^2 + C \implies C = \dfrac12

Then the particular solution is

\dfrac12 y^2 = \dfrac12 x^2 + \dfrac12

y^2 = x^2 + 1

y = \pm\sqrt{x^2 + 1}

and because y(0)=-1, we take the negative solution to accommodate this initial value.

\boxed{y(x) = -\sqrt{x^2+1}}

7 0
2 years ago
Inverse of f(x) = x2 − 9
luda_lava [24]
To find the inverse of a function, replace every x in the equation with a y, and replace every y in the equation with an x:

x = y^{2} - 9

Add 9 to both sides:

y^{2} = x + 9

Square root both sides to get y by itself:

y = \sqrt{x+9}

This equation can be simplified by taking the square root of 9 out of the root:

\sqrt{9} = 3
y = 3 + \sqrt{x}

The inverse of this function is y = 3 + √x.
7 0
3 years ago
Read 2 more answers
Help i dont understand this
Igoryamba
I’m not extremely good at math but I think you need to make -x-y=1 into y=mx+b form and then solve your new equation!
4 0
3 years ago
60=F(15) <br><br> Please solve this and show work, please
Lubov Fominskaja [6]

Answer:

f = 4

Step-by-step explanation:

60= 15f

divide both side by 15

f = 4

3 0
3 years ago
Consider the following equation. f(x, y) = y3/x, P(1, 2), u = 1 3 2i + 5 j (a) Find the gradient of f. ∇f(x, y) = Correct: Your
BaLLatris [955]

f(x,y)=\dfrac{y^3}x

a. The gradient is

\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath

\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}

b. The gradient at point P(1, 2) is

\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}

c. The derivative of f at P in the direction of \vec u is

D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}

It looks like

\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath

so that

\|\vec u\|=\sqrt{\left(\dfrac{13}2\right)^2+5^2}=\dfrac{\sqrt{269}}2

Then

D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}

\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}

7 0
3 years ago
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