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Natasha2012 [34]
3 years ago
11

Happy Birthday Scorpios!!

Mathematics
2 answers:
Sonja [21]3 years ago
8 0

Answer:

HAPPY B-DAY :))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))

Step-by-step explanation:

juin [17]3 years ago
6 0

Answer:

Happy Birthday Scorpios!!

Step-by-step explanation:

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You know that 12x=60, and that x=5. Which property of equality would you use to conclude that 12•5=60?
alexandr402 [8]

ANSWER

The substitution property of equality.

EXPLANATION

We know that:

12x = 60

and that

x = 5

Since x=5 or 5=x , we can substitute x for 5 wherever we see x in the given equation.

When we do that, we obtain;

12 \bullet \: 5 = 60

This is what we refer to as the substitution property of equality.

5 0
3 years ago
Rain is falling steadily in Seattle, Washington. After 6 hours, 4 inches of rain has fallen. How long will it take for 1 inch of
seraphim [82]
You do 6/4, so the answer should be: 1.5 hours!
5 0
2 years ago
If 5x2+4x+xy=3 and y(3)=−18, what is an equation of the tangent line to the graph at the point (3,−18)
Paha777 [63]
I believe the answer is:

y+18 = (-52/3)(x-3) in point slope form
 
Let me know if you need an explanation. Hope this helps.
8 0
3 years ago
Please help me <br> math math math
olga nikolaevna [1]
Your answer A12/A6= A6
4 0
2 years ago
Can we obtain a diagonal matrix by multiplying two non-diagonal matrices? give an example
polet [3.4K]
Yes, we can obtain a diagonal matrix by multiplying two non diagonal matrix.

Consider the matrix multiplication below

\left[\begin{array}{cc}a&b\\c&d\end{array}\right]   \left[\begin{array}{cc}e&f\\g&h\end{array}\right] =  \left[\begin{array}{cc}a e+b g&a f+b h\\c e+d g&c f+d h\end{array}\right]

For the product to be a diagonal matrix,

a f + b h = 0 ⇒ a f = -b h
and c e + d g = 0 ⇒ c e = -d g

Consider the following sets of values

a=1, \ \ b=2, \ \ c=3, \ \ d = 4, \ \ e=\frac{1}{3}, \ \ f=-1, \ \ g=-\frac{1}{4}, \ \ h=\frac{1}{2}

The the matrix product becomes:

\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}\frac{1}{3}&-1\\-\frac{1}{4}&\frac{1}{2}\end{array}\right] = \left[\begin{array}{cc}\frac{1}{3}-\frac{1}{2}&-1+1\\1-1&-3+2\end{array}\right]= \left[\begin{array}{cc}-\frac{1}{6}&0\\0&-1\end{array}\right]

Thus, as can be seen we can obtain a diagonal matrix that is a product of non diagonal matrices.
8 0
3 years ago
Read 2 more answers
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