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3 years ago

13

The longest side of a triangle is 80cm more than the second and 130cm more than the shortest side and its perimeter is 540cm. fi

nd its area
help help

1 answer:

Phoenix [80]3 years ago

5 0

Answer:

Area of triangle1/2b*h Now you can do

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Please answer my question, this is the third time i've asked. (75 Points)

natta225 [31]

For A it is 6240 / 20% for B it is 6240 x 15%

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A photograph has a length that is 2 inches longer than its width, x. So its area is given by the expression x(x + 2) square inch

Tatiana [17]

Width = 6 inches I hope this helps!

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3 years ago

2/3x=10 thankyou for helping me

Novay_Z [31]

(2/3)x = 10
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PLEASE HELP<br> 5.<br> If t= -3, then 3t^2 +5t +6 equals<br> A. -36<br> B. -6<br> c. 6 <br> D. 18

Nataliya [291]

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D

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3 years ago

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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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3 years ago

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