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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Murljashka [212]

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3 years ago

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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 Citizen ticket.
Call 'S' the price of the Student ticket.

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

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

Those are your two simultaneous equations. Now the idea is to multiply or divide each side of one equation in such a way that when you add or subtract it from the other equation, one of the variables will become a zero quantity ... you'll be left with an equation in one variable, which you can easily solve. THEN, knowing the value of one variable, you can put it back into one of the original equations,and find the value of the other variable.

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
32 C + 12 S = 296

Now, subtract the lower equation from the upper one, and you get . . .

- 22 C + 0 = - 88

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

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 : S = $ 14.00 .

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:

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