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Aleksandr [31]
3 years ago
11

(9.3×10 to the sixth power) plus (1.8×10 to the fourth power)

Mathematics
1 answer:
balandron [24]3 years ago
8 0

Answer:

9.318  ×  10^ 6

Step-by-step explanation:

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the  10. If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.

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Choose the equation below that represents the line passing through the point (1, −4) with a slope of one half
murzikaleks [220]
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2 years ago
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AB = 3 + x
ICE Princess25 [194]

The given quadrilateral ABCD is a parallelogram since the opposite sides are of same length AB and DC is 4 and AD and BC is 2.

<u>Step-by-step explanation</u>:

ABCD is a quadrilateral with their opposite sides are congruent (equal).

The both pairs of opposite sides are given as AB = 3 + x , DC = 4x , AD = y + 1 , BC = 2y.

  • AB and DC are opposite sides and have same measure of length.
  • AD and BC are opposite sides and have same measure of length.

<u>To find the length of AB and DC :</u>

AB = DC

3 + x = 4x

Keep x terms on one side and constant on other side.

3 = 4x - x

3 = 3x

x = 1

Substiute x=1 in AB and DC,

AB = 3+1 = 4

DC = 4(1) = 4

<u>To find the length of AD and BC :</u>

AD = BC

y + 1 = 2y

Keep y terms on one side and constant on other side.

2y-y = 1

y = 1

Substiute y=1 in AD and BC,

AD = 1+1 = 2

BC = 2(1) = 2

Therefore, the opposite sides are of same length AB and DC is 4 and AD and BC is 2. The given quadrilateral ABCD is a parallelogram.

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3 years ago
What's the additive identity of 9
Zigmanuir [339]
9? Thats what it is,
5 0
3 years ago
What is the simplified form of x+3/4+x+2/4
Savatey [412]
Hey there! :D

Since all the terms are being added together, and there is a common denominator with the fractions, we can just add them.

x+x= 2x  3/4+2/4= 5/4

2x+5/4

2x+1.25<== simplest form

I hope this helps!
~kaikers
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3 years ago
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