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Paha777 [63]
3 years ago
14

How do u solve this equation

Mathematics
1 answer:
galina1969 [7]3 years ago
8 0
Hi

So first off, you don't want a negative 6/5y

So you move it to the other side to make it positive, now your equation is

24=6/5y

Now divide both sides by 6/5 to get y by itself

y=20

Hope this helps




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4 years ago
Need help ASAP
joja [24]

Part (1) : The solution is 729

Part (2): The solution is $\frac{1}{16 x^{8}}$

Part (3): The solution is $\frac{2 x^{2}}{3 y z^{7}}$

Explanation:

Part (1): The expression is 3^{2} \cdot3^{4}

Applying the exponent rule, $a^{b} \cdot a^{c}=a^{b+c}$, we get,

$3^{2} \cdot 3^{4}=3^{2+4}$

Adding the exponent, we get,

3^{2} \cdot3^{4}=3^6=729

Thus, the simplified value of the expression is 729

Part (2): The expression is $\left(2 x^{2}\right)^{-4}$

Applying the exponent rule, $a^{-b}=\frac{1}{a^{b}}$, we have,

$\left(2 x^{2}\right)^{-4}=\frac{1}{\left(2 x^{2}\right)^{4}}$

Simplifying the expression, we have,

\frac{1}{2^4x^8}

Thus, we have,

$\frac{1}{16 x^{8}}$

Thus, the value of the expression is $\frac{1}{16 x^{8}}$

Part (3): The expression is $\frac{2 x^{4} y^{-4} z^{-3}}{3 x^{2} y^{-3} z^{4}}$

Applying the exponent rule, $\frac{x^{a}}{x^{b}}=x^{a-b}$, we have,

\frac{2x^{4-2}y^{-4+3}z^{-3-4}}{3}

Adding the powers, we get,

\frac{2x^{2}y^{-1}z^{-7}}{3}

Applying the exponent rule, $a^{-b}=\frac{1}{a^{b}}$, we have,

$\frac{2 x^{2}}{3 y z^{7}}$

Thus, the value of the expression is $\frac{2 x^{2}}{3 y z^{7}}$

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4 years ago
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4s/4=p/4

s=1/4p

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3 years ago
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ICE Princess25 [194]

Answer:

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=>

f(i) = i³- 2i²

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:)

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3 years ago
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that's some weird logic but ok
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