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TEA [102]
2 years ago
9

Which is the exponential form of log9 5 = y

Mathematics
1 answer:
nasty-shy [4]2 years ago
3 0

Answer:

Step-by-step explanation:

\log _9\left(5\right)=y\\y=\log _9\left(5\right)

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What's 5/6 as a decimal rounded to the nearest hundredth
zaharov [31]
Hello there!

the answer is 8.33

Hope I helped!

Let me know if you need anything else!

~ Zoe
4 0
2 years ago
What values of c and d make the equation true? RootIndex 3 StartRoot 162 x Superscript c Baseline y Superscript 5 Baseline EndRo
Reil [10]

Answer:

<em>c=6, d=2</em>

Step-by-step explanation:

<em>Equations </em>

We must find the values of c and d that make the below equation be true

\sqrt[3]{162x^cy^5}=3x^2y \sqrt[3]{6y^d}

Let's cube both sides of the equation:

\left (\sqrt[3]{162x^cy^5}\right )^3=\left (3x^2y \sqrt[3]{6y^d}\right)^3

The left side just simplifies the cubic root with the cube:

162x^cy^5=\left (3x^2y \sqrt[3]{6y^d}\right)^3

On the right side, we'll simplify the cubic root where possible and power what's outside of the root:

162x^cy^5=3^3x^6y^3 (6y^d)

Simplifying

x^cy^5=x^6y^{3+d}

Equating the powers of x and y separately we find

c=6

5=3+d

d=2

The values are

\boxed{c=6,d=2}

3 0
2 years ago
Read 2 more answers
Bryan wants to ship three medium-sized boxes. Two companies offer different rates.
nikitadnepr [17]
I think th answer is a
5 0
2 years ago
Read 2 more answers
Arrange the entries of matrix A in increasing order of their cofactors values
givi [52]

To find the cofactor of

A=\left[\begin{array}{ccc}7&5&3\\-7&4&-1\\-8&2&1\end{array}\right]

We cross out the Row and columns of the respective entries and find the determinant of the remaining 2\times 2 matrix with the alternating signs.


Ac_{11}=\left|\begin{array}{ccc}4&-1\\2&1\end{array}\right|


Ac_{11}=4\times 1- -1\times 2


Ac_{11}=4+ 2

Ac_{11}=6




Ac_{12}=-\left|\begin{array}{ccc}-7&-1\\-8&1\end{array}\right|


Ac_{12}=-(-7\times 1- -1\times -8)


Ac_{12}=-(-7- 8)

Ac_{12}=15




Ac_{21}=-\left|\begin{array}{ccc}5&3\\2&1\end{array}\right|


Ac_{21}=-(5\times 1- 3\times 2)


Ac_{21}=-(5-6)


Ac_{21}=1







A_c{23}=-\left|\begin{array}{ccc}7&5\\-8&2\end{array}\right|


Ac_{23}=-(7\times 2 -8\times 5)


Ac_{23}=-(14-40)


Ac_{23}=26




A_c{31}=\left|\begin{array}{ccc}5&3\\4&-1\end{array}\right|


Ac_{31}=5\times -1 -4\times 3


Ac_{31}=-5-12


Ac_{31}=-17


A_c{33}=\left|\begin{array}{ccc}7&5\\-7&4\end{array}\right|


Ac_{33}=7\times 4- -7\times 5


Ac_{33}=28+35


Ac_{33}=63


Therefore in increasing order, we have;

Ac_{31}=-17,Ac_{21}=1,Ac_{11}=6,Ac_{23}=26,Ac_{12}=15, Ac_{33}=63



7 0
2 years ago
Read 2 more answers
If the mean of a symmetric distribution is 170, which of these values could be the median of the distribution?
tankabanditka [31]
The answer is a.

Symmetric distribution have the same mean and median.
6 0
2 years ago
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