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NeX [460]
2 years ago
8

Mrs. Rhodes is building the ramp shown below and she wants to paint it blue. How much paint will she need to cover the entire su

rface?
Mathematics
1 answer:
Mademuasel [1]2 years ago
4 0

Answer:

2/3 of paint I believe because of the amount shown

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I needa know how to do this
Elena L [17]

Answer:

So I (3,0) becomes I' (1,3)

H (5,2) H' (1,1)

G (2,4) G' (2,1)

Step-by-step explanation:

So let's start with I, just count how many times you move to the left then how many times you move up.

UR WELCOME BABEEEE~ <3

5 0
2 years ago
Write the equation of the line that passes through the points (-7, 7) and (-6, -5).
defon

Answer:

y = -12x - 77

Step-by-step explanation:

The standard form of the equation is expressed as y = mx+b

m is the slope

Given the coordinate (-7, 7) and (-6, -5).

m = -5-7/-6+7

m = -12/1

m = -12

Get the intercept b

Substitute m= -12 and (-7 7) into y = mx+b

7 = -12(-7) + b

7 = 84 + b

b = 7 - 84

b = -77

Get the required equation

y = mx+b

y = -12x +(-77)

y = -12x - 77

This gives the required equation

4 0
2 years ago
Line BD is tangent to circle O at C, Arch AEC = 299, and ACE = 93. Find Angle DCE.
VladimirAG [237]
In triangle ACE,
we know C=93,E can be calulated by using arch angle AEC...what ever that is....,using this we get  A=180-(E+93)

So, by alternate segment theorem, DCE= A.

thats all i can say.
7 0
2 years ago
2+2 so you can get 40 pts I need to just write 20 characters so... ndds8bdsudbsd8s shdds8sjd sudh89hd
Zanzabum

Answer:

2+2 is 4

Step-by-step explanation:

You can imagine 2 cookies and then you get 2 more which is 4.

4 0
2 years ago
The plane x+y+2z=8 intersects the paraboloid z=x2+y2 in an ellipse. Find the points on this ellipse that are nearest to and fart
DiKsa [7]

Answer:

The minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

Step-by-step explanation:

Here, the two constraints are

g (x, y, z) = x + y + 2z − 8  

and  

h (x, y, z) = x ² + y² − z.

Any critical  point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we  actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from (x, y, z) to the origin is given by

√((x − 0)² + (y − 0)² + (z − 0)² ).

This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema  of the square of the distance. Thus, our objective function is

f(x, y, z) = x ² + y ² + z ²

and

∇f = (2x, 2y, 2z )

λ∇g = (λ, λ, 2λ)

µ∇h = (2µx, 2µy, −µ)

Thus the system we need to solve for (x, y, z) is

                           2x = λ + 2µx                         (1)

                           2y = λ + 2µy                       (2)

                           2z = 2λ − µ                          (3)

                           x + y + 2z = 8                      (4)

                           x ² + y ² − z = 0                     (5)

Subtracting (2) from (1) and factoring gives

                     2 (x − y) = 2µ (x − y)

so µ = 1  whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0  into (3) gives us  2z = −1  and thus z = − 1 /2 . Subtituting z = − 1 /2  into (4) and (5) gives us

                            x + y − 9 = 0

                         x ² + y ² +  1 /2  = 0

however, x ² + y ² +  1 /2  = 0  has no solution. Thus we must have x = y.

Since we now know x = y, (4) and (5) become

2x + 2z = 8

2x  ² − z = 0

so

z = 4 − x

z = 2x²

Combining these together gives us  2x²  = 4 − x , so

2x²  + x − 4 = 0 which has solutions

x =  (-1+√33)/4

and

x = -(1+√33)/4.

Further substitution yeilds the critical points  

((-1+√33)/4; (-1+√33)/4; (17-√33)/4)   and

(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).

Substituting these into our  objective function gives us

f((-1+√33)/4; (-1+√33)/4; (17-√33)/4) = (195-19√33)/8

f(-(1+√33)/4; - (1+√33)/4; (17+√33)/4) = (195+19√33)/8

Thus minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

4 0
2 years ago
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