How to add and subtract positive and negative fractions

Mathematics

2 answers:

Dvinal [7]3 years ago

5 0

Rules for adding: if the bases are the same you can add them, and subtracting happens in a fraction

Alenkinab [10]3 years ago

5 0

1. create a common denominator.
2. multiply the numerator by the same number . Next , you multiply the top numbers (numerators)by the same number as the denominators .
3. add (or subtract ) the numerators.
4. simplify if necessary.

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Convert the polar coordinates (-3, -60°) to Cartesian coordinates.

notsponge [240]

Answer:

(1.5,-2.6)

Step-by-step explanation:

Given the polar coordinates (-3,60°).

Let our Cartesian coordinates be (x,y)

#We know that when converting the rectangular coordinates (x,y) to polar (r,θ), then:

$r=\sqrt{x^2+y^2}\\\\\therefore r^2=x^2+y^2\\\\\theta=tan^{-1}(y/x)\\\therefore tan \theta=y/x$

#Using the illustration above, we can express our polar coordinates as:

$-3=\sqrt{x^2+y^2}\\\\-60\textdegree=tan^{-1}(y/x}$

#Solve simultaneously to solve for x and y:

$(-3)^2=x^2+y^2\ \ \ \ \ \ \ \ \ \ i\\\\tan(-0\texdegree)=y/x\ \ \ \ \ \ \ \ ...ii\\\\y=x\ tan(-60\textdegree)\ \ \ \ \ \ \ ...iii\\\\\#substitute\ y \ in\ i\\\\(-3)^2=x^2+(x \ tan (-60\textdegree))^2\\\\9=x^2+3x^2\\\\x=\sqrt{9/4}=1.5\\\\y=1.5\ tan(-60\textdegree)=-2.5981\approx-2.6$

Hence, the Cartesian coordinates are (1.5,-2.6)

6 0

3 years ago

Can someone explain how I multiply both sides ?

stellarik [79]

Answer:

x=3

Step-by-step explanation:

Given,

1 = 1/(x-2)

Multiply both sides by (x-2),

1*(x-2) = 1/(x-2) * (x-2)

x-2 = 1

Adding 2 on both sides,

x-2+2 = 1+2

x = 3

6 0

3 years ago

❊ Simplify :

Nataliya [291]

To simplify the given expression .

$\red{\frak{Given}}\Bigg\{ \sf \dfrac{x - 1}{ {x}^{2} - 3x + 2} + \dfrac{x - 2}{ {x}^{2} - 5x + 6 } + \dfrac{x - 5}{ {x}^{2} - 8x + 15 }$

$\rule{200}4$

$\sf\longrightarrow \small \dfrac{x - 1}{ {x}^{2} - 3x + 2} + \dfrac{x - 2}{ {x}^{2} - 5x + 6 } + \dfrac{x - 5}{ {x}^{2} - 8x + 15 } \\\\\\\sf\longrightarrow \small \dfrac{ x-1}{x^2-x -2x +2} +\dfrac{ x-2}{x^2-3x-2x+6} +\dfrac{ x -5}{x^2-5x -3x + 15 } \\\\\\\sf\longrightarrow\small \dfrac{ x -1}{ x ( x - 1) -2(x-1) } +\dfrac{ x-2}{x ( x -3) -2( x -3)} +\dfrac{ x -5}{ x(x-5) -3( x -5) } \\\\\\\sf\longrightarrow \small \dfrac{ x -1}{ ( x-2) (x-1) } +\dfrac{ x-2}{( x -2)(x-3) } +\dfrac{ x -5}{ (x-3)(x-5) } \\\\\\\sf\longrightarrow\small \dfrac{ 1}{ x-2} +\dfrac{ 1}{ x -3} +\dfrac{1}{ x -3} \\\\\\\sf\longrightarrow \small \dfrac{1}{x-2} +\dfrac{2}{x-3} \\\\\\\sf\longrightarrow \small \dfrac{ x-3 +2(x-2)}{ ( x -3)(x-2) } \\\\\\\sf\longrightarrow \small \dfrac{ x - 3 +2x -4 }{ (x-3)(x-2) } \\\\\\\sf\longrightarrow \underset{\blue{\sf Required \ Answer }}{\underbrace{\boxed{\pink{\frak{ \dfrac{ 3x -7}{ ( x -2)(x-3) } }}}}}$