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1 year ago

PLEASEE HELPP. Im rlly struggling please help if you can. with steps

Mathematics

2 answers:

Tom [10]1 year ago

8 0

X= 3 ,y= 6 ,z= -1
^Answer^
Solve for the first variable in one of the equations, then substitute the result into the other equation.
Point Form: (3,6,−1)

Nadusha1986 [10]1 year ago

7 0

Answer:

lol

Step-by-step explanation:

L O LDJDDHDHDHDHFFHFVDVDVDBSSBDBDBDBD

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3 years ago

Can somebody step by step explain how to solve this problem??<br><br> (3x^2y)(-7xy^3)

charle [14.2K]

Answer:

$\boxed{\bold{-21x^3y^4}}$

Explanation:

[ Step One ] Remove Parenthesis: (a) = a

$\bold{-3x^2y\cdot \:7xy^3}$

[ Step Two ] Multiply: $\bold{3\cdot \:7=21}$

$\bold{-21x^2xy^3y}$

[ Step Three ] Apply Exponent Rule $\bold{\:a^b\cdot \:a^c=a^{b+c}: \ x^2x=\:x^{2+1}=\:x^3}$

$\bold{-21x^3y^3y}$

[ Step Four ] Apply Exponent Rule: $\bold{y^3y=\:y^{1+3}=\:y^4}$

$\bold{-21x^3y^4}$

[ $\boxed{\bold{Final \ Answer}}$ ]

➤ $\bold{-21x^3y^4}$

$\boxed{\bold{Mordancy}}$

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3 years ago

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Oxana [17]

Answer:

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Step-by-step explanation:

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3 years ago

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Which of the following interpretations for the given expression is correct? (29) the cube of the difference of the square of 5 t

masya89 [10]

Answer:

Step-by-step explanation:

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3 years ago

Are the functions f(x) = (x^2-1)/(x-1) and g(x)= x+1 equal for all x?

Vinvika [58]

$\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-------------------------------\\\\$

$\bf f(x)=\cfrac{x^2-1}{x-1}\implies f(x)=\cfrac{x^2-1^2}{x-1}\implies f(x)=\cfrac{(\underline{x-1})(x+1)}{\underline{x-1}}
\\\\\\
f(x)=x+1\qquad \qquad \qquad \qquad \qquad g(x)=x+1\\\\
-------------------------------\\\\
\textit{they're, kinda, except that, when x = 1}
\\\\\\
g(x)=(1)+1\implies g(x)=2
\\\\\\
f(x)=\cfrac{(1)^2-1}{(1)-1}\implies f(x)=\cfrac{0}{0}\impliedby und efined$

7 0

3 years ago