Is -94 higher or -5 please answer this is for homework sksksksksk and I oop

1 answer:

4 0

If you want to know which is greater, it has to be the negative number that is closer to zero so in this case it would be -5

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Sugar was traditionally produced and sold as sugarloaves, which are cones of sugar wrapped in paper. Find the total surface area

Zolol [24]

Answer:

A = 326,73 cm²

Step-by-step explanation:

The area of a circular cone is area of the base (A₁ ) plus area lateral ( area of a circular sector of radius the slant height )

Then we proceed to calculate the area of the base A₁

diameter of circular base is 8 cm then the radius is 4 cm and the area is:

A₁ = π*r² = 3,14* (4)²

A₁ = 3,1416*16 = 50,2656 cm²

Now Lateral area of the cone (A₂) is equal to the area of a circular sector with radius the slant height. We will calculate it, taken into account that this circular sector is part of a circle of radius the slant height.

Between the area of circular sector with radius the slant height and the area of the circle with the same radius there is a linear relation. That is we can calculate area of a circular sector by rule of three as follows:

The length of the circular sector is the length of the circle of the base of the cone, that is:

L = 2*π*(4)

L = 2*3,1416*4

L = 25,1328 cm

Then we have a circular sector of length 25,1328 cm

The area of the circle of radius 22 cm is:

A(c) = π*(22)² ⇒ A(c) = 1520,5344 cm²

And the length of this circle is:

L(c) = 2*π*(22) ⇒ 138,2304 cm

Then we apply a rule of three

For a length of 138,2304 cm ⇒⇒⇒ (area) 1520,5344 cm²

Then for a length of 25,1328 cm ⇒⇒⇒(area) A₂ (??)

Therefore:

A₂ = (25,1328)*1520,5344)/ 138,2304

A₂ = 276,4608 cm²

Then total area of the cone is:

A = A₁ + A₂

A = 50,2656 +276,4608

A = 326,7264 cm²

Round answer A = 326,73 cm²

8 0

2 years ago

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What is the value of (7 + 2i)(3 - i)?

Harrizon [31]

There are two ways to work this out: normal variables or using "imaginary" numbers.

Normal variables:
$(7+2i)(3-i)\\(7*3)+[7*(-i)]+(3*2i)+[2i*(-i)]\\21-7i+6i-2i^{2}\\\\21-i-2i^{2}$

Imaginary numbers:
Using the result from earlier:
$21-i-2i^{2}$
Now since $i = \sqrt{-1}$ , then the expression becomes:
$21-i-2i^{2}\\21-i-2(\sqrt{-1})^{2}\\21-i+(-2)(-1)\\21-i+2\\\\23-i$