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Maru [420]
3 years ago
10

What is the rectangular form of 6(cos(225)+ i sin(225))?

Mathematics
1 answer:
LekaFEV [45]3 years ago
8 0

Answer: A is the answer

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Solve 2 (3d + 3) = 7 + 6d.
Leni [432]

First distribute in the 2 on the left side to get 6d+6=7+6d

Then, you would combine the variables to be on one side of the equation by doing -6d. However, this would cancel out the d by leaving you with 0d+6=7. That is the same thing as saying 6=7.

The answer to this equation is no solution because 6 is NOT equal to 7.

4 0
3 years ago
Read 2 more answers
Jem boy wants to make his 8-meter square pool into a rectangular one by increasing its length by 2m and decreasing its width by
butalik [34]
Side = 8m
He increases length by 2m and reduces width by 2m
(a.) New dimensions will be -
Length = 10m and Width = 6m
(b.) Area = length x breadth (or width)
 here we will use the special product (x+y) (x-y)
length = (x+y) and width = (x-y)
area = (8+2) x (8-2)
            8^(2) - 2^(2) 
            = 60
3 0
3 years ago
The price of gas has been increasing over the last month. Renee believes there is a positive correlation between the number of p
snow_tiger [21]

Step-by-step explanation:

To find a rat of change (ROC) you take the two numbers and the two prices and subtract them to make a ratio<u>:</u>

<u>_</u><u>_</u><u>_</u><u>_</u><u>6</u><u>-</u><u>3</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u> = <u>_</u><u>_</u><u>_</u><u>3</u><u>_</u><u>_</u><u>_</u>

$2.56-$2.44 = $2.12

The reduce to the lowest number (the 3 to a 1)

<u>_</u><u>_</u><u>3</u><u>_</u><u>_</u><u>_</u> = <u>_</u><u>_</u><u>_</u><u>1</u><u>_</u><u>_</u>

$2.12 $0.71

The ROC is $0.71

8 0
3 years ago
Read 2 more answers
During a guided tour of Sindrow Castle, a group of tourists climbed to the top of Fineview Tower.
Finger [1]

Answer:

C

Step-by-step explanation:

7 0
2 years ago
How do you factor this? x^{2} + 5 x- 9 = ?
Oksanka [162]
x^2+5x-9\\\\a=1;\ b=5;\ c=-9\\\Delta=b^2-4ac\\\\\Delta=5^2-4\cdot1\cdot(-9)=25+36=61 > 0\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{61}\\\\x_1=\dfrac{-5-\sqrt{61}}{2\cdot1}=\dfrac{-5-\sqrt{61}}{2}\\\\x_2=\dfrac{-5+\sqrt{61}}{2\cdot1}=\dfrac{-5+\sqrt{61}}{2}\\\\x^2+5x-9=\left(x-\dfrac{-5-\sqrt{61}}{2}\right)\left(x-\dfrac{-5+\sqrt{61}}{2}\right)
6 0
3 years ago
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