Help for 1,2,3,4 please

Mathematics

1 answer:

ikadub [295]3 years ago

5 0

In every case, you're finding the surface area of a rectangular prism. That area is the sum of the areas of the 6 rectangular faces. Since opposite faces have the same area, the formula can be written

... S = 2(LW +WH +HL)

The number of multiplications can be reduced if you rearrange the formula to

... S = 2(LW +H(L +W))

where L, W, and H are the length, width, and height of the prism. (It does not matter which dimension gets what name, as long as you use the same number for the same variable in the formula.)

When you're evaluating this formula over and over for diffferent sets of numbers, it is convenient to let a calculator or spreadsheet program do it for you.

1. S = 2((5 cm)(5 cm) +(5 cm)(5 cm +5 cm)) = 2(25 cm² +(5 cm)(10 cm))

... = 2(25 cm² + 50 cm²) = 150 cm²

2. S = 2(12·6 + 2(12+6)) mm² = 2(72 +36) mm² = 216 mm²

3. S = 2(11·6 + 4(11 +6)) ft² = 2·134 ft² = 264 ft²

4. S = 2(10·4 +3(10 +4)) in² = 164 in²

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What is the value of the expression i 0 × i 1 × i 2 × i 3 × i 4?

astraxan [27]

ANSWER

The value of the expression is
$- 1$

EXPLANATION

Method 1: Rewrite as product of
${i}^{2}$

The expression given to us is,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4}$

We use the fact that
${i}^{2} = - 1$
to simplify the above expression.

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0} \times {i}^{1} \times {i}^{3} \times {i}^{2} \times {i}^{4}$

This implies,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2}$

We substitute to obtain,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = 1\times - 1 \times - 1 \times - 1\times - 1 \times - 1$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = 1\times 1 \times 1 \times - 1 = - 1$

Method 2: Use indices to solve.

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0 + 1 + 2 + 3 + 4}$

This implies that,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{10}$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = ( {{i}^{2}} )^{5}$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = ( - 1 )^{5} = - 1$

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Mars2501 [29]

Answer:

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Step-by-step explanation:

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