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2 years ago

Kalo kita toxic di malam lailatul qadar itu toxicnya sama kaya 1000 bulan gak

Mathematics

1 answer:

VARVARA [1.3K]2 years ago

8 0

I don’t know understand but yeah okay

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Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st

Naily [24]

Answer:

$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

Step-by-step explanation:

I will try to give as many details as possible.

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$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

$\boxed{a^{0} = 1, a\neq 0 }$

$\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

Note

$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

Note

$\boxed{\dfrac{1}{\dfrac{1}{a} } = a}$

$9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)$

Once

$9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6$

$\boxed{(a \cdot b)^n=a^n \cdot b^n}$

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27}$

We have

$\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)$

Also, once

$\boxed{\dfrac{c^a}{c^b}=c^{a-b}}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27$

$30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3$

$2^{\frac{802}{27}} \cdot 3^9$

4 0

2 years ago

How could I get rid of the 3 shown in equation 1?

Tems11 [23]

Answer:

Step-by-step explanation:

3 log 2+log (x-4)=log 2³+log (x-4)=log 8+log (x-4)=log 8(x-4)

8 0

2 years ago

What is the inverse of the function f(x) = 2x + 1?

BaLLatris [955]

Answer:

$f^{-1} = \frac{x-1}{2}$

Step-by-step explanation:

$f(x) = 2x+1$

Replace it with y

$y = 2x+1$

Exchange the values of x and y

$x = 2y+1$

Solve for y

$x = 2y+1$

Subtracting 1 from both sides

$2y = x-1$

Dividing both sides by 2

$y = \frac{x-1}{2}$

Replace it by $f^{-1}$

So,

$f^{-1} = \frac{x-1}{2}$

3 0

3 years ago

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Write an equation reprezeting x axis

Rudiy27

Answer:

y=0

Step-by-step explanation:

All y-values along the x-axis are equal to 0.

3 0

2 years ago

Find the value of x.

FinnZ [79.3K]

Answer:

x = 5

Step-by-step explanation:

14x + 22x = 180

36x = 180

x = 5

7 0

2 years ago

Read 2 more answers