Please help ive been trying and i just cant get it right

Mathematics

1 answer:

avanturin [10]3 years ago

5 0

The correct answer is c. :)

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Simplify this step by step<br> <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B-5jk%7D%7B35j%5E2k%5E2%7D%20" id="TexFormula1" tit

VikaD [51]

$\frac { -5jk }{ 35{ j }^{ 2 }{ k }^{ 2 } } \\ \\ =-\frac { 5 }{ 35 } \cdot \frac { j }{ { j }^{ 2 } } \cdot \frac { k }{ { k }^{ 2 } } \\ \\ =-\frac { 1 }{ 7 } \cdot \frac { 1 }{ j } \cdot \frac { 1 }{ k } \\ \\ =-\frac { 1 }{ 7jk }$

Remember that:

$\frac { j }{ { j }^{ 2 } } =\frac { 1 }{ j } \cdot \frac { j }{ j } =\frac { 1 }{ j } \cdot 1=\frac { 1 }{ j } \\ \\ \frac { k }{ { k }^{ 2 } } =\frac { 1 }{ k } \cdot \frac { k }{ k } =\frac { 1 }{ k } \cdot 1=\frac { 1 }{ k }$

8 0

3 years ago

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What are the real zeros of the function g(x)=x^3+2x^2-x-2

Gre4nikov [31]

$g(x)=x^3+2x^2-x-2\\\\\text{The zeros:}\\\\g(x)=0\to x^3+2x^2-x-2=0\\\\x^2(x+2)-1(x+2)=0\\\\(x+2)(x^2-1)=0\iff x+2=0\ \vee\ x^2-1=0\\\\x+2=0\ \ \ |-2\\x=-2\\\\x^2-1=0\ \ \ |+1\\x^2=1\to x=\pm\sqrt{1}\to x=-1\ \vee\ x=1\\\\\text{Answer:}\ x=-2\ or\ x=-1\ or\ x=1$

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3 years ago

How will you know that the equation is a quadratic

prohojiy [21]

Answer:

A quadratic condition is a condition of the subsequent degree, which means it contains, in any event, one term that is squared. The standard structure is ax² + bx + c = 0 with a, b, and c being constants, or mathematical coefficients, and x is an obscure variable.

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3 years ago

Can u put 0.5,0.41 and 3/5 in least to greatest

baherus [9]

0.41, 0.5, 3/5 is the order

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3 years ago

Y=2x-11 and -x -y=-4 solve by substitution

IceJOKER [234]

Answer:

(5, -1)

Step-by-step explanation:

8 0

3 years ago