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aleksandr82 [10.1K]
2 years ago
8

Help me please. I will mark Brainliest

Mathematics
2 answers:
Firdavs [7]2 years ago
7 0

Answer:

true

Step-by-step explanation:

lora16 [44]2 years ago
5 0

Answer:

this is false

Step-by-step explanation:

it's false because 8 is not the same value as 1/8

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A: A circle has a circumference of 45 cm. What is the BEST estimate of the circle’s diameter?
dedylja [7]

Answer:

14

Step-by-step explanation:

d=c/pi

45/pi=14

6 0
3 years ago
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If Erin built a rectangular prism shaped trough for her horses to eat from the trough is 3 m long 1 m wide and 1 m high hay for
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It would cost $33.00 hope that helps
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3 years ago
Please help me and show your work
morpeh [17]

Answer:

Step-by-step explanation:

y = 3x + 4

Plugin x = 3 in the equation,

y = 3*3 + 4

 = 9 + 4

y = 13

Plugin x = 4 in equation,

y = 3*4 + 4

  = 12 + 4

y = 16

Plugin x = 5 in the equation,

y = 3*5 + 4

  = 15 + 4

y = 19

2) y =x - 7

Plugin x = 10 in the equation

y = 10 - 7

y = 3

Plugin x = 15 in the equation

y = 15 - 7

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y = 20 - 7

y = 13

6 0
2 years ago
'find the equation of the line shown':
insens350 [35]

Answer:

y = x + 6

Step-by-step explanation:

The y-intercept is (0, 6).  This is where the line crosses the y-axis.

The slope is 4/4, or 1:  m = 1

The equation of the line is therefore y = x + 6

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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