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TiliK225 [7]
3 years ago
12

Which of the following statements is not true?

Mathematics
2 answers:
Nat2105 [25]3 years ago
6 0
X5y5z5 = xyz5 is not true
Ilya [14]3 years ago
5 0
Hey there!

To find if the first two answers are true, all you need to do is count the number of variables of each type in each equation. This number will be your exponent, since it's the number of times you will multiply that variable by itself. 

For your first answer, there are 6 x's. This means that when you multiply the x's together, it will equal x⁶. This is correct. 

For your second answer, there are 7 A's and 5 B's. This will equal A⁷B⁵. This is correct. 

The last two answers are a bit different. In order to simplify an equation like the ones you've been given, you'd need to put all of your terms in parenthesis in order to say that all of the terms within them must be multiplied by what is outside of the parenthesis. This means that:

(bc)⁶ = (b)⁶(c⁶) = b⁶c⁶, which is correct

xyz⁵ = xyz⁵ ≠ x⁵y⁵z⁵, which is not correct

Your answer will be your fourth option. Hope this helped you out! :-)
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Answer:I believe 37.25

Step-by-step explanation:

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3 years ago
Can you explain how to do 13 please
Zigmanuir [339]
I must confess that I was about to pass this question by, but I was captured by the respectful and dignified way in which you asked for help. 

A careful reading of the problem gives you two equations in two unknowns, which you can then solve as simultaneous equations.  Here's how it looks:

Call 'C' the price of the senior <u>C</u>itizen ticket.
Call 'S' the price of the <u>S</u>tudent ticket.

On the first night . . .    10 C  +  12 S  =  208

On the second night . . .  8C  +  3 S  =  74

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This all sounds more complicated than it is.   Here's how it goes:
We have . . .

10 C  +  12 S  =  208  and
8C  +  3 S  =  74

I'm going to multiply each side of the second equation by  4, and then write it under the first one:

10 C  +  12 S  =  208
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- 22 C  +  0  =   - 88

Divide each side of this one by  -22  and you have    <em>C = $4.00</em> .

THAT's what you need, to blow the whole problem wide open.  Knowing
the value of 'C', let's substitute it into the equation for the first night:

10 C  +  12 S  =  208

10(4)  +  12 S  =  208

40  +  12 S  =  208

Subtract  40  from each side :    12 S  =  168

Divide each side by  12 :    <em>S  =</em><em>  $ 14.00 </em>.

Finally, as we look over our results, and see that Students have to pay  $14  to see the show but Seniors can get in for only  $4 , we reflect on this ... or at least I do ... and realize that getting old is not necessarily all bad.
3 0
3 years ago
(−t 4 −5t 3 −10t 2 )+(9t 3 +3t 2 −1)
klemol [59]

Answer:

\left(-t^4-5t^3-10t^2\right)+\left(9t^3+3t^2-1\right)=-t^4+4t^3-7t^2-1

Step-by-step explanation:

Given the expression

\left(-t^4\:-5t^3\:-10t^2\:\right)+\left(9t^3\:+3t^2\:-1\right)

Remove parentheses:  (a)=a

=-t^4-5t^3-10t^2+9t^3+3t^2-1

Group like terms

=-t^4-5t^3+9t^3-10t^2+3t^2-1

Add similar elements        

=-t^4-5t^3+9t^3-7t^2-1         ∵ -10t^2+3t^2=-7t^2

Add similar elements        

=-t^4+4t^3-7t^2-1                  ∵  -5t^3+9t^3=4t^3

Thus, the equivalent expression in simplified form:

\left(-t^4-5t^3-10t^2\right)+\left(9t^3+3t^2-1\right)=-t^4+4t^3-7t^2-1

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