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3 years ago

The log of x cubed times y squared simplified

Mathematics

1 answer:

tresset_1 [31]3 years ago

4 0

Answer:

$log(x^{3}y^{2}) = 3 log x+2 log y$

Step-by-step explanation:

Step 1:-

using logarithmic formula $log(ab)=log a+log b$

so given $log(x^{3} y^{2} ) = log (x^{3} )+log(y^{2} )$

now simplify

= $3 log x+2 log y$

Answer:-

$log(x^{3}y^{2})= [tex]3 log x+2 log y$

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X=6. y=5. 12×+3y please help me

IgorLugansk [536]

Answer:

Your answer: 87

Step-by-step explanation:

Since you are given what both variables are in the equation, this problem requires you to plug it in.

12(6)+3(5)= ?

You can either do this in a calculator, or do this one step at a time.

12(6)=72

3(5)=15

72+15=87

7 0

3 years ago

6x + 8=5x + 12 i don’t get this

NeX [460]

Answer: x = 4

Step-by-step explanation:

6x + 8 = 5x + 12

-8 -8

6x = 5x + 4

-5x -5x

x = 4

5 0

2 years ago

Help me with this pleaseee

Galina-37 [17]

Answer:

option c

Step-by-step explanation:

Volume = Length * Width * Height

this is a pretty easy problem cuz all u need to do is find which one of the options has a product of 729.

option a) 7 * 8 * 12

well 7 * 8 * 12 equals 672, so that cannot be the box

option b) 8*9*11

8*9*11 = 792. cant be the box cuz 792 does not equal 729

option c) 9*9*9

9*9*9=729

option c is the box cuz 729 = 729

5 0

2 years ago

Use the bionomial theorem to write the binomial expansion

MaRussiya [10]

Answer:

$\left(\frac{x}{2} + 3 y\right)^{4}=\frac{x^{4}}{16} + \frac{3}{2} x^{3} y + \frac{27}{2} x^{2} y^{2} + 54 x y^{3} + 81 y^{4}$

Step-by-step explanation:

$\left(\frac{1}{2}x+3y \right)^4=\left(\frac{x}{2}+3y \right)^4\\$

Binomial Expansion Formula:

$(a+b)^n=\sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$ , also $\binom{n}{k}=\frac{n!}{(n-k)!k!}$

We have to solve $\left(\frac{x}{2} + 3 y\right)^{4}=\sum_{k=0}^{4} \binom{4}{k} \left(3 y\right)^{4-k} \left(\frac{x}{2}\right)^k$

Now we should calculate for $k=0, k=1, k=2, k=3 \text{ and } k =4;$

First, for $k=0$

$\binom{4}{0} \left(3 y\right)^{4-0} \left(\frac{x}{2}\right)^{0}=\frac{4!}{(4-0)! 0!}\left(3 y\right)^{4} \left(\frac{x}{2}\right)^{0}=\frac{4!}{4!}(81y^4)\cdot 1 =81 y^{4}$

It is the same procedure for the other:

For $k=1$

$\binom{4}{1} \left(3 y\right)^{4-1} \left(\frac{x}{2}\right)^{1}=54 x y^{3}$

For $k=2$

$\binom{4}{2} \left(3 y\right)^{4-2} \left(\frac{x}{2}\right)^{2}=\frac{27}{2} x^{2} y^{2}$

For $k=3$

$\binom{4}{3} \left(3 y\right)^{4-3} \left(\frac{x}{2}\right)^{3}=\frac{3}{2} x^{3} y$

For $k=4$

$\binom{4}{4} \left(3 y\right)^{4-4} \left(\frac{x}{2}\right)^{4}=\frac{x^{4}}{16}$

You can perform the calculations, I will not type everything.

The answer is the sum of elements calculated.

Just organizing:

$\left(\frac{x}{2} + 3 y\right)^{4}=\frac{x^{4}}{16} + \frac{3}{2} x^{3} y + \frac{27}{2} x^{2} y^{2} + 54 x y^{3} + 81 y^{4}$

8 0

3 years ago

Read 2 more answers

Can someone help geometry assp

melomori [17]

Answer:

24°

Step-by-step explanation:

66°+b=90°(right angle)

or,b=90°-66°

or,b=24°

hope this helps you

4 0

3 years ago