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pshichka [43]
2 years ago
8

Simplify: -75 sq rt over sq rt of 25

Mathematics
1 answer:
ivann1987 [24]2 years ago
8 0

Answer:

\frac{ \sqrt{ - 75} }{ \sqrt{25} }  \\  \\  \frac{ - 5625}{625}  \\  \\   - 9

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-2.1b +5.3= b -- 3.1b + 5.3<br><br> does is have one solution or infinitely many solutions ? Why ?
pychu [463]

Answer: infinitely many solutions.

Step-by-step explanation:

Ok, our equation is:

-2.1*b + 5.3 = b - 3.1*b + 5.3

now, simplifyng the right side, we have:

b - 3.1*b + 5.3 = (1 - 3.1)*b + 5.3 = -2.1*b + 5.3

Then our initial expression is:

-2.1*b + 5.3 = -2.1*b + 5.3

So in both sides of the equality we have the exact same thing, so this is a trivial equality.

This means that the equality will remain true for any value of b, which means that we have infinitely many solutions.

6 0
3 years ago
Before the 1936 US presidential election, a popular poll predicted that Alfred Landon would comfortably win against Franklin D.
Vinvika [58]

Answer:

wrong subject

Step-by-step explanation:

3 0
2 years ago
One third of a number t is equal to 7
Vikentia [17]
If one third of t is 7 than t is equal to 21
3 0
2 years ago
Read 2 more answers
Find f(-10) F(x) =7x-5<br><br> A= -75<br> B=-65<br> C=65<br> D=90
STALIN [3.7K]

Answer:

-75

Step-by-step explanation:

f(-10) = 7(-10) - 5

f(-10) = -70 - 5

f(-10) = -75

8 0
3 years ago
Hard math question!
luda_lava [24]
First, let me show you some notation.

To show a matrix is an inverse of another matrix, we write A^{-1}

-1 is not an exponent. It just shows that a matrix is an inverse of another matrix.

For a 2x2 matrix, we can get the inverse by first making b and c negatives and swap the positions of a and d.

Then multiply each entry in the matrix by 1 divided by the determinant.

\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]^{-1} = &#10;  \frac{1}{ad - bc}\left[\begin{array}{ccc}d&{-b}\\{-c}&a\end{array}\right] =  \\  \\ \\ \left[\begin{array}{ccc}d(\frac{1}{ad-bc})&{-b}(\frac{1}{ad-bc}) \\ {-c}(\frac{1}{ad-bc}) &a(\frac{1}{ad-bc}) \end{array}\right]

I hope this helped!
8 0
2 years ago
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