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Triss [41]
3 years ago
12

SIMPLIFY 81 TO THE 1/3 POWER

Mathematics
2 answers:
Ne4ueva [31]3 years ago
3 0
81^1/3. We know that 81 is 3^4, meaning 3 x 3 x 3 x 3 is 81. So (3^4)^1/3 is 3^4/3, which is 3 * (cubed root of 3).
solniwko [45]3 years ago
3 0
Okay lets see 81^1/3 means that you are multiplying by a third. but you can also turn it into a division problem as simple as putting 81 in the place of the 1. Now the problem looks like this: 81/3. so if you divide 81 by 3 you get 27. check the ,mat by doing it backwards. multiply 27 by 3 and see what you get. it should be 81.
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Cindy's Ceramics engaged in a like-kind exchange that resulted in a $3,000 gain. In addition to the like-kind property, Cindy al
irinina [24]

Answer:

True

Step-by-step explanation:

$3000

-lesser of realized gain or boot

8 0
3 years ago
Find the value or measure. Assume all lines that appear to be tangent are tangent. X=
aev [14]

According to the secant-tangent theorem, we have the following expression:

(x+3)^2=10.8(19.2+10.8)

Now, we solve for <em>x</em>.

\begin{gathered} x^2+6x+9=10.8(30) \\ x^2+6x+9=324 \\ x^2+6x+9-324=0 \\ x^2+6x-315=0 \end{gathered}

Then, we use the quadratic formula:

x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}

Where a = 1, b = 6, and c = -315.

\begin{gathered} x_{1,2}=\frac{-6\pm\sqrt[]{6^2-4\cdot1\cdot(-315)}}{2\cdot1} \\ x_{1,2}=\frac{-6\pm\sqrt[]{36+1260}}{2}=\frac{-6\pm\sqrt[]{1296}}{2} \\ x_{1,2}=\frac{-6\pm36}{2} \\ x_1=\frac{-6+36}{2}=\frac{30}{2}=15 \\ x_2=\frac{-6-36}{2}=\frac{-42}{2}=-21 \end{gathered}<h2>Hence, the answer is 15 because lengths can't be negative.</h2>
7 0
1 year ago
what is 372x19+89 WANT FREE POINTS? COME HERE SHHHH! I HAVE TO ASK A QUESTION SO IT DOESNT GET DELETED! LET ME KNOW IF YOU WANT
soldi70 [24.7K]

Answer:

7157

Step-by-step explanation:

Answer is 7157

3 0
4 years ago
Read 2 more answers
Type the correct answer in each box. Use numerals instead of words. For this question, any answer that is not a whole number sho
egoroff_w [7]

Answer:

Step-by-step explanation:

Since we are not given the value of P, |Q|, R and S, we can as well assume values for them for the sake of this question.

Let P = 5, |Q| = 6, R =7 |S| = 2

Note that since Q and S are in modulus sign, they can return both positive and negative values.

P+Q = 5 + 6 (note that the positive value of Q is used since we need the greatest value of P+Q)

P+Q = 11

Hence the greatest value of P+Q is 11

For the least value of P+Q, we will use the negative value of Q as shown

P+Q = 5+(-6)

P+Q = 5-6

P+Q = -1

Hence the least value of P+Q is -1

Similarly:

R+S = 7 + 2 (note that the positive value of S is used since we need the greatest value of R+S)

R+S = 9

Hence the greatest value of R+S is 9

For the least value of R+S, we will use the negative value of S as shown

R+S = 5+(-2)

R+S = 5-3

R+S = 2

Hence the least value of P+Q is 2

NOTE THAT THIS ARE ASSUMED VALUES. ALL YOU NEED IS TO PLUG IN THE VALUES OF P, Q and R THAT YOU HAVE IN CASE THE VALUES DIFFERS.

5 0
3 years ago
9t - 6 - 6t = 6<br><br> show work
ICE Princess25 [194]
9t-6-6t=6 -- add 6 to both sides
9t-6t=12 -- combine like terms
3t=12 -- divide by 3

t=4
7 0
3 years ago
Read 2 more answers
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