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Ber [7]
3 years ago
10

I need to know how to solve it

Mathematics
2 answers:
Wewaii [24]3 years ago
8 0
I think first you should 7-42 then you will get -35 which will be -7x-35. I hope that I was helpful





Snowcat [4.5K]3 years ago
3 0
-7x > -42. So you divide on both sides by -7 and that gives you x > 6
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Help please, will give lots of points!
Troyanec [42]

Step-by-step explanation:

roots as exponents are fractions

if there is no root number next to the radical sign its assumed that its 1/2

3√2 is 2^(1/3)

√3 is 3^(1/2)

4√3 is 3^(1/4)

5√2 is 2^(1/5)

8 0
2 years ago
Tyrone is buying candy by the pound to stuff into a piñata. Her purchased 14 pounds of candy for $12. How much is the cost of on
Deffense [45]
I believe the answer is


26

sorry if wrong


if not an answer choice then 168.
4 0
3 years ago
Solve equation (n-1)(n+6)(n+5)=0
Vika [28.1K]
Answer:
1, -6 and -5 

Explanation:
To solve the equation means to get the values of n which satisfy the given equation.
The equation given is:
(n-1) (n+6) (n+5) = 0
This equation will be true if any of the terms (brackets) is equal to zero.
This means that:
either n-1 = 0 .............> n = 1
or n + 6 = 0 ................> n = -6
or n + 5 = 0 ................> n = -5

Hope this helps :)
5 0
3 years ago
Read 2 more answers
Add the following values.<br> (2.1 × 1016) + (3.2 × 1015)
qwelly [4]

Answer:

5381.6

Step-by-step explanation:

(2.1 × 1016) + (3.2 × 1015)

= 5381.6

8 0
2 years ago
Plz help<br>urgent !!!!<br>will give the brainliest !!​
Mkey [24]

Answer:

X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}

Step-by-step explanation:

The question we have at hand is, in other words,

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix} - where we have to solve for the value of X

If we have to isolate X here, then we would have to take the inverse of the following matrix ...

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix} ... so that it should be as follows ... \begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}

Therefore, we can conclude that the equation as to solve for " X " will be the following,

X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} - First find the 2 x 2 matrix inverse of the first portion,

\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1} = \frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}= \frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix} = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} ,

X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix} - Simplifying this, we should get ...

\begin{bmatrix}1&3\\ 2&4\end{bmatrix} ... which is our solution.

7 0
3 years ago
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