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Scrat [10]
3 years ago
10

Evaluate A2 for A = -3. -9 -6 6 9

Mathematics
2 answers:
elena55 [62]3 years ago
8 0
Ok so basically it is saying substitue when a equals all of those numbers,so basically I will show you.

a (times) 2 = ?   so substitute one of the numbers for a which would be like this its basically multiplication so, (-3) (times) 2 = -6 so that solves one part do the same with the other numbers and that will give you all the answers like so,

(-9) (times) 2 = -18   (-6) (times) 2 = -12    (6) (times) 2 = 12 and finally the last one is (9) (times) 2 = 18 which are all correct answers.

so basically, your answers are gonna be -6,-18,-12,12,18 which are or should be all correct
schepotkina [342]3 years ago
5 0
Evaluate A² for A = -3.

(-3)² = (-3) * (-3) = 9

Your answer is 9.

If the question is, however, evaluate A2, which is 2A, for A = -3, then the answer is:

2A = 2(-3) = -6.
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108.15 all together

Step-by-step explanation:


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4 years ago
Carol rolled a dice 150 times she rolled a 6 29 times use carols data to estimate the probability of rolling a six with this dic
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3 0
3 years ago
Order the set of integers from least to greatest 5, -3, -12, 9
Sauron [17]

Answer:

-12, -3, 5, 9

Step-by-step explanation:

ok. So the numbers are 5, -3, -12, and 9.

In these numbers, there are 2 negatives. They are -12 and -3.

In negatives, the bigger the number is, the number is smaller.

So we first put

-12, -3 as our 2 first. Then the other ones are 5 and 9.

and you obviously know that 5 is less than 9. try it out. 9-5 =4 which is positive

So then you put 5, 9

And our full answer is

-12, -3, 5, 9

7 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=x%5Ea%3Dy%5Eb%3Dz%5Ec%20%5C%5Cxyz%3D1%5C%5Cab%2Bbc%2Bca%3D%3F" id="TexFormula1" title="x^a=y^b
SashulF [63]

Answer:

<em>ab+bc+ca=0</em>

Step-by-step explanation:

We are given:

x^a=y^b=z^c

xyz=1

Separating equations:

x^a=z^c

y^b=z^c

Solving the first equation for x:

\displaystyle x=z^\frac{c}{a}

Solving the second equation for y:

\displaystyle y=z^\frac{c}{b}

Substituting in

xyz=1:

\displaystyle z^\frac{c}{a}z^\frac{c}{b}z=1

Adding the exponents:

\displaystyle z^{\frac{c}{a}+\frac{c}{b}+1}=1

Since 1=z^0:

\displaystyle z^{\frac{c}{a}+\frac{c}{b}+1}=z^0

The base is z on both sides so we get rid of them:

\displaystyle \frac{c}{a}+\frac{c}{b}+1=0

Multiplying by ab:

bc+ac+ab=0

Reordering:

ab+bc+ca=0

5 0
3 years ago
Help me :D i appreciate anyone who can!
dmitriy555 [2]

Answer:

I don’t understand what do we have to do?

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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