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balandron [24]
3 years ago
8

Simplify the following expressions without using a calculator

Mathematics
1 answer:
wolverine [178]3 years ago
6 0

Answer:

13x^3y^4 i am not sure

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Solve the equation identify extraneous solutions
PIT_PIT [208]

Answer:

Step-by-step explanation:

The goal to solving any equation is to have x = {something}.  That means we need to get the x out from underneath that radical.  It's a square root, so we can "undo" it by squaring.  Square both sides because this is an equation.  Squaring both sides gives you

x^2=-3x+40

Get everything on one side of the equals sign and set the quadratic equal to 0:

x^2+3x-40=0

Throw this into the quadratic formula to get that the solutions are x = 5 and -8.  We need to see if only one works, both work, or neither work in the original equation.

Does 5=\sqrt{-3(5)+40}?

5=\sqrt{-15+40} and

5=\sqrt{25}

and 5 = 5.  So 5 works.  Let's try -8 now:

-8=\sqrt{-3(-8)+40} and

-8=\sqrt{24+40} so

-8=\sqrt{64}

-8 = 8?  No it doesn't.  So only 5 works. Your choice is the third one down.

6 0
3 years ago
The diagram shows the net of a juice box. The box is a rectangular prism. What is the surface area of the juice box?
LenaWriter [7]

Answer:269.6 square centimeters

Step-by-step explanation:

5 0
3 years ago
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
3 years ago
Read 2 more answers
Can someone please help me I’m stuck
olga_2 [115]

Answer:

25

Step-by-step explanation:

You have to do base x height divided by 2

50/2

8 0
3 years ago
What’s 9 * 9 * 9 * 9 as an exponent?
CaHeK987 [17]

Answer:

9^{4}

Step-by-step explanation:

9^{4} means 9*9*9*9

7 0
3 years ago
Read 2 more answers
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