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Dahasolnce [82]

3 years ago

9

Luis found a new text messaging plan which will charge him $2.00 for 80 messages. Using this plan, how much would he pay for 900

text messages in one month

1 answer:

Andreas93 [3]3 years ago

4 0

Let x be # of text messages
y = 2(x/80)
Plug 900 in to x,
y = 2(900/80)
y= 22.50

Luis would have to pay $22.50

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Write the equation of a line that is perpendicular to y=−1 and that passes through the point (8,-4)

Vera_Pavlovna [14]

Answer:

perp. slope: 1

y + 4 = x - 8

y = x - 12

5 0

3 years ago

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,

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

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

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

8 0

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6 0

3 years ago

The hypotenuse of a right triangle is 10cm long. The longer leg is 2cm longer than the shorter leg. Find the side lengths of the

Crazy boy [7]

We need Pythagoras theorem here
a^2+b^2 = c^2
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Let S=shorter leg, in cm, then longer leg=S+2 cm
use Pythagoras theorem
S^2+(S+2)^2 = (10 cm)^2
expand (S+2)^2
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simplify and form standard form of quadratic
S^2+2S-48=0
Solve by factoring
(S+8)(S-6) = 0 means (S+8)=0, S=-8
or (S-6)=0, S=6
Reject nengative root, so
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Hypotenuse (given) = 10 cm

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Do 3 divided by 53 to find the cost of one ticket and then multiply the that number with 7

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