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Daniel [21]
3 years ago
11

Y=mx+b solve for x.

Mathematics
2 answers:
Rus_ich [418]3 years ago
7 0
You can't solve for x. Y=mx+b is slope intercept form and is used to graph equations
Svetlanka [38]3 years ago
4 0
<span>Equation of a straight line with slope m and y intercept b is: y=mx+b. To find the x intercept,solve this equation by taking y =0. So, it will become mx +b=0, mx=-c, and finally x=-c/m. I hope this helped!</span>
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In the above figure, AE = 2, CE = 3, and DE = 4. What is the length of BE?
Gre4nikov [31]
BE= 6. I had this question not too long ago. Hope this helps. :)
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Jon borrows $6.50 from his friends to pay for snacks , and $11.75 from his sister to go to the movies. How much money does he ne
Naddik [55]

Answer:

$18.25

Step-by-step explanation:

5 0
3 years ago
Can someone explain this differentiation question to me? I can differentiate but then I'm not sure what I am doing
weeeeeb [17]

note that gradient = \frac{dy}{dx} at x = a

calculate \frac{dy}{dx} for each pair of functions and compare gradient

(a)

\frac{dy}{dx} = 2x and \frac{dy}{dx} = - 1

at x = 4 : gradient = 8 and - 1 : 8 > - 1

(b)

\frac{dy}{dx} = 2x + 3 and \frac{dy}{dx} = - 2

at x = 2 : gradient = 7 and - 2 and 7 > - 2

(c)

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(d)

\frac{dy}{dx} = 6x - 5 and \frac{dy}{dx} = 2x - 2

at x = - 1 : gradient = - 11 and - 4 and - 4 > - 11

(e)

y = √x = x^{\frac{1}{2} }

\frac{dy}{dx} = 1/(2√x) and \frac{dy}{dx} = 2

at x = 9 : gradient = \frac{1}{6} and 2 and 2 > \frac{1}{6}


4 0
3 years ago
A television costs $270.25 including 15% tax. How much of cost is the tax
My name is Ann [436]

Answer:

229.7125

Step-by-step explanation:

4 0
3 years ago
Find the point on the parabola y^2 = 4x that is closest to the point (2, 8).
guapka [62]

Answer:

(4, 4)

Step-by-step explanation:

There are a couple of ways to go at this:

  1. Write an expression for the distance from a point on the parabola to the given point, then differentiate that and set the derivative to zero.
  2. Find the equation of a normal line to the parabola that goes through the given point.

1. The distance formula tells us for some point (x, y) on the parabola, the distance d satisfies ...

... d² = (x -2)² +(y -8)² . . . . . . . the y in this equation is a function of x

Differentiating with respect to x and setting dd/dx=0, we have ...

... 2d(dd/dx) = 0 = 2(x -2) +2(y -8)(dy/dx)

We can factor 2 from this to get

... 0 = x -2 +(y -8)(dy/dx)

Differentiating the parabola's equation, we find ...

... 2y(dy/dx) = 4

... dy/dx = 2/y

Substituting for x (=y²/4) and dy/dx into our derivative equation above, we get

... 0 = y²/4 -2 +(y -8)(2/y) = y²/4 -16/y

... 64 = y³ . . . . . . multiply by 4y, add 64

... 4 = y . . . . . . . . cube root

... y²/4 = 16/4 = x = 4

_____

2. The derivative above tells us the slope at point (x, y) on the parabola is ...

... dy/dx = 2/y

Then the slope of the normal line at that point is ...

... -1/(dy/dx) = -y/2

The normal line through the point (2, 8) will have equation (in point-slope form) ...

... y - 8 = (-y/2)(x -2)

Substituting for x using the equation of the parabola, we get

... y - 8 = (-y/2)(y²/4 -2)

Multiplying by 8 gives ...

... 8y -64 = -y³ +8y

... y³ = 64 . . . . subtract 8y, multiply by -1

... y = 4 . . . . . . cube root

... x = y²/4 = 4

The point on the parabola that is closest to the point (2, 8) is (4, 4).

4 0
3 years ago
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