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gtnhenbr [62]
2 years ago
7

Which is a factor of each term of the polynomial? (3d2 – 10d)

Mathematics
1 answer:
melomori [17]2 years ago
5 0
(3d^2 - 10d) = d(3d - 10).

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PLEASE ANSWER THANK YOU SO MUCH GUYS:)
babunello [35]

Answer:

  • We khow that the company charges 0.75 dollars to each ship
  • so c= 0.75* p
  • when the price is reduced the equaton will be :
  • c= 0.7*p

4 0
3 years ago
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Can someone pleasw helppppppp meee
Sergio [31]
It would be 2,5, and 6. The mode means the number that is repeating the most. Since 2, 5, and 6 all appear 3 times, that is the mode
6 0
2 years ago
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What is the probability of randomly selecting an odd number from the numbers 1 - 15?
Darya [45]

Answer:

7/15 Hope it helps

Step-by-step explanation:

5 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
2 years ago
PLS HELP THIS IS HARD ANYONE PLS
Valentin [98]

This question is very oddly worded.  The domain is the set of x-values, but this is a set of (x,y) ordered pairs.

I'm reading this question as "Here's a function, { (1,5), (2,1), (-1,-7) }.  If this is reflected over the x-axis, what's the range?"

Assuming that is the question that is meant to be asked, reflecting a function over the x-axis will just change the signs of the y-values.

    (1,5) -> (1,–5)

    (2,1) -> (2,–1)

    (-1,-7) -> (-1,+7)

I'd pick the third option.

6 0
3 years ago
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