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3 years ago

10

(4j3 - 7j + 2s3 ) - ( - 12j3 - 8j )

1 answer:

OLga [1]3 years ago

7 0

Answer:

$-8j^3+j+2s^3$

Step-by-step explanation:

The trick here is to consider $j^3$ , $j$ , and $s^3$ as separate variables.

$(4j^3-7j+2s^3)-(-12j^3-8j)$

$4j^3-7j+2s^3+12j^3+8j$

$4j^3+12j^3-7j+8j+2s^3$

$16j^3+j+2s^3$

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Can someone help me? I’ll reward points + brainalist

salantis [7]

Answer:

C

Step-by-step explanation:

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3 years ago

Multiply 3.76 X.2<br> Show your work

WITCHER [35]

Answer:

7.52

Step-by-step explanation:

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I hope this helps!

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3 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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3 years ago

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Lunna [17]

Answer:

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2 years ago

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Dafna1 [17]

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3 years ago