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3 years ago

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Use the formula SA=6s2 , where SA is the surface area and s is the edge length of the cube, to solve this problem. Eliz must cov

er a cube-shaped box with giftwrap. The edge length of the box is 6 1/2 inches. What is the surface area of the box?

1 answer:

Vika [28.1K]3 years ago

4 0

This it's my answer with the formula I used shown. Hope this helps. God bless

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Simplify the expression -2+6.45z-6+(-3.25z)

erik [133]

Answer:

-8 + 3.2z

Step-by-step explanation:

when there is a "+" in front of an expression in parentheses, the expression remains the same
-2 + 6.45z - 6 - 3.25z
calculate the difference

-8 + 6.45z - 3.25z

collect like terms

-8 + 3.2z

8 0

3 years ago

Neil took his family of five out to dinner at HogTastic Pit BBQ. They all ordered the buffet and drink at $12.80 each. Neil left

Reika [66]

The answer would be 73.60 dollars hope this helps!!!

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3 years ago

gogolik [260]

Answer:

The statements about arcs and angles that are true in the figure are;

1) ∠EFD ≅ ∠EGD

2) $\overline{ED}\cong \overline{FD}$

3) mFD = 120°

Step-by-step explanation:

1) ∠ECD + ∠CEG + ∠CDG + ∠GDE = 360° (Sum of interior angle of a quadrilateral)

∠CEG = ∠CDG = 90° (Given)

∠GDE = 60° (Given)

∴ ∠ECD = 360° - (∠CEG + ∠CDG + ∠GDE)

∠ECD = 360° - (90° + 90° + 60°) = 120°

∠ECD = 2 × ∠EFD (Angle subtended is twice the angle subtended at the circumference)

120° = 2 × ∠EFD

∠EFD = 120°/2 = 60°

∠EFD ≅ ∠EGD

∠ECD = 120°

∠EGD = 60°

∴∠EGD ≠ ∠ECD

2) Given that arc mEF ≅ arc mFD

Therefore, ΔECF and ΔDCF are isosceles triangles having two sides (radii EC and CF in ΔECF and radii EF and CD in ΔDCF

∠ECF = mEF = mFD = ∠DCF (Given)

∴ ΔECF ≅ ΔDCF (Side Angle Side, SAS, rule of congruency)

$\\ \overline{EF}\cong \overline{FD}$ (Corresponding Parts of Congruent Triangles are Congruent, CPCTC)

∠FED ≅ ∠FDE (base angles of isosceles triangle)

∠FED + ∠FDE + ∠EFD = 180° (sum of interior angles of a triangle)

∠FED + ∠FDE = 180° - ∠EFD = 180° - 60° = 120°

∠FED + ∠FDE = 120° = ∠FED + ∠FED (substitution)

2 × ∠FED = 120°

∠FED = 120°/2 = 60° = ∠FDE

∴ ∠FED = ∠FDE = ∠EFD = 60°

ΔEFD is an equilateral triangle as all interior angles are equal

$\\ \overline{EF}\cong \overline{FD}\cong \overline{ED}$ (definition of equilateral triangle)

$\overline{ED}\cong \overline{FD}$

3) Having that ∠EFD = 60° and ∠CFE = ∠CFD (CPCTC)

Where, ∠EFD = ∠CFE + ∠CFD (Angle addition)

60° = ∠CFE + ∠CFD = ∠CFE + ∠CFE (substitution)

60° = 2 × ∠CFE

∠CFE =60°/2 = 30° = ∠CFD

$\overline{CF}\cong \overline{CD}$ (radii of the same circle)

ΔFCD is an isosceles triangle (definition)

∠CFD ≅ ∠CDF (base angles of isosceles ΔFCD)

∠CFD + ∠CDF + ∠DCF = 180°

∠DCF = 180° - (∠CFD + ∠CDF) = 180° - (30° + 30°) = 120°

mFD = ∠DCF (definition)

mFD = 120°.

5 0

3 years ago

Which is the equation for the nth term of the geometric sequence-2, 8, -32, ... ?

Cerrena [4.2K]

Answer:

54

Step-by-step explanation:

4 0

3 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

3 years ago