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Gala2k [10]
3 years ago
5

Add the following decimals 0.654+0.45

Mathematics
2 answers:
anzhelika [568]3 years ago
6 0

When adding decimals you would line up the decimal points and fill blank spots with zeros.

so...

  0.654

+ 0.450

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

  <u>1.104</u>.

Mice21 [21]3 years ago
3 0
When adding up decimals you have to aline the decimals and fill in the blank spots with zeros. Then you add like you normally would.
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For c = d(3.14), what is c when d = 100 show work
Katyanochek1 [597]

Answer:314

Step-by-step explanation: 3.14 x 100 = 314

Move the decimal two times to the right because there is two zeros

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2 years ago
For the feasibility region shown below, find the maximum value of the function P=2x+3y.
timofeeve [1]
Corner points in this graph are: ( 0,0 ) ( 0,8 ) ( 5,6 ) and ( 8, 0 ).
If we plug those values in : P = 2 x + 3 y
P ( 0,0 )= 0
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2 years ago
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Emma read the statement “the quotient of six and a number, x, is the same as negative two times the difference of x and 4.5.” Sh
Angelina_Jolie [31]
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2 years ago
Two points in the Cartesian plane are A(2.00 m, −4.00 m) and B(−3.00 m, 3.00 m). Find the distance between them and their polar
Brrunno [24]

The distance between the two points is d=8.6m

The polar coordinate of A is \left(4.47,296.57\right)

The polar coordinate of B is \left(4.24,135\right)

Explanation:

The two points are A(2,-4) and B(-3,3)

The distance between two points is given by,

d=\sqrt{(2+3)^{2}+(-4-3)^{2}}\\d=\sqrt{(5)^{2}+(-7)^{2}}\\d=\sqrt{25+49}\\d=8.6

Thus, the distance between the two points is d=8.6m

The polar coordinates of A can be written as (Distance, tan^{-1} \frac{y}{x} )

Distance = \sqrt{x^{2} +y^{2} }

Substituting A(2,-4), we get,

Distance = \sqrt{2^{2} +(-4)^{2} }=\sqrt{4+16 }=4.47

tan^{-1} \frac{y}{x} =tan^{-1} \frac{-4}{2}=-63.43

To make the angle positive, let us add 360,

\theta=360-63.43=296.57

The polar coordinate of A is \left(4.47,296.57\right)

Similarly, The polar coordinate of B can be written as (Distance, tan^{-1} \frac{y}{x} )

Distance = \sqrt{x^{2} +y^{2} }

Substituting B(-3,3), we get,

Distance = \sqrt{(3)^{2}+(3)^{2}}=4.24

tan^{-1} \frac{y}{x} =tan^{-1} \frac{3}{-3}=-45

To make the angle positive, let us add 360,

\theta=180-45=135^{\circ}

The polar coordinate of B is \left(4.24,135\right)

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