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professor190 [17]
4 years ago
6

Based on the graph, which inequality is correct for a number that is to the right of -4

Mathematics
1 answer:
Anon25 [30]4 years ago
3 0

2 > -4

2 is greater than -4

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Help me find the solution set for 8x-3=2(x-1/2)
daser333 [38]
8x -3 = 2(x-1/2)
⇒ 8x -3= 2x -2*(1/2)
⇒ 8x -3 = 2x -1
⇒ 8x -2x= -1+ 3
⇒ 6x = 2
⇒ x= 2/6
⇒ x= 1/3

The final answer is x= 1/3~
6 0
3 years ago
Ill mark brainlist plss help
mojhsa [17]

Answer:

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Step-by-step explanation:

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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
<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%3F%7D%7B5%7D%20%2A%20%5Cfrac%7B10%7D%7B8%7D%20%3D1.5%5C%5C" id="TexFormula1" title=
Serggg [28]

Answer:

\boxed{\sf \huge \boxed{ \sf ?} = 6}

Step-by-step explanation:

\sf Solve  \: for \: \boxed{ \sf ?} :  \\  \sf \implies \frac{\boxed{ \sf ?}}{5}  \times  \frac{10}{8}  = 1.5 \\  \\  \sf  \implies \frac{\boxed{ \sf ?} \times 10}{5 \times 8}  = 1.5 \\  \\  \sf \frac{10}{5}  =  \frac{ \cancel{5} \times 2}{ \cancel{5}}  = 2 :  \\  \sf \implies  \frac{ \boxed{ \sf 2} \times \boxed{ \sf ?} }{8}  = 1.5 \\  \\  \sf \frac{2}{8}  =  \frac{ \cancel{2}}{ \cancel{2}  \times 4}  =  \frac{1}{4}  :  \\  \sf \implies \frac{\boxed{ \sf ?}}{ \boxed{ \sf 4}}  = 1.5 \\  \\  \sf Multiply  \: both  \: sides \:  of \:  \frac{\boxed{ \sf ?}}{4}  = 1.5 \: by \: 4 :  \\  \sf \implies \frac{4 \times \boxed{ \sf ?}}{4}  = 4 \times 1.5 \\  \\  \sf \frac{4 \times \boxed{ \sf ?}}{4} =  \frac{ \cancel{4}}{ \cancel{4}}  \times \boxed{ \sf ?} = \boxed{ \sf ?} :  \\  \sf  \implies \boxed{\boxed{ \sf ?}} = 4 \times 1.5 \\  \\  \sf 4 \times 1.5 = 6 :  \\  \sf \implies \boxed{ \sf ?} = 6

4 0
3 years ago
Which number would you divide the numerator and the denominator of the first fraction to yield the second fraction?
blondinia [14]

Answer:

the number I would to divide the numerator and denominator is 5/5

8 0
3 years ago
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