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vfiekz [6]
2 years ago
9

Please let me know if this is correct and if not what is the correct answer. Thanks

Mathematics
2 answers:
vlada-n [284]2 years ago
6 0
There’s nothing on the picture
mart [117]2 years ago
3 0

Answer:

is it about your future profession or your dream

Step-by-step explanation:

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Which statement is true about the expression 7(2x 5)A. The expression has two factorsB. The expression has three factors.C. The
agasfer [191]
I think that the answer is a because you only have to terms after you distribute hope this helps you
7 0
3 years ago
Write an expression for the sequence of operations described below.
MariettaO [177]
Given question ,

Raise v to the 5th power then triple the result

The expression is as follows :

v ^{5}

Tripling , (v^5)^{3}
6 0
1 year ago
The function y=5x+10 represents the total cost y of a group of x members to go paintballing.
never [62]

Answer:

The independent variable (x) represents the members to go paintballing.

The dependent variable (y) represents the total cost.

Step-by-step explanation:

The y is always dependent on the x. The number of people attending affects the total cost of the trip. The more people attend, the more it costs.

3 0
3 years ago
PLEASE HELP ME! 100 POINTS! ;( question 10, 11, 12, 13, 14, 15. I have a f in this class and could use your help a lot! I will l
MatroZZZ [7]

There are a lot of problems here so I'll try to be brief with each one so I don't add a lot of clutter

Problem 10) This is <u>arithmetic</u> because we subtract 6 from each term to get the next. This is the same as adding on -6. The common difference is d = -6. The first term is a = 1

Problem 11) <u>Neither</u>. The distance from 3 to 3/2 = 1.5 is 1.5 units, but the distance from 1.5 to 1 is 0.5 units. This implies there is no common difference value. The sequence is not arithmetic because of this. It's also not geometric either. To go from 3 to 3/2, we multiply by 1/2. But then going from 3/2 to 1, we multiply by 2/3. There is no common ratio r value.

Problem 12) This sequence is <u>geometric</u>. We divide each term by 3 to get the next, or put another way, we multiply each term by 1/3 to get the next term. The common ratio is r = 1/3. The first term is a = 108.

Problem 13) <u>Neither</u>. This is because going from term to term, we do not add the same amount each time. Example: from -2 to 4 we add 6, but then from 4 to -6 we add -10. So the sequence is not arithmetic. It's also not geometric either because we dont multiply by the same term each time. Eg: from -2 to 4, we multiply by -2; but from 4 to -6 we multiply by -1.5

Problem 14) The next three terms are: <u>1, 1/4, 1/16</u>. This is found by multiplying each term by the common ratio r = 1/4, or you can think of dividing each term by 4. To get this common ratio, pick any term you want and divide it by its previous term. Example: term2/term1 = 16/64 = 1/4 = common ratio.

Problem 15) The next three terms are: <u>-432, 2592, -15552</u>. You multiply each term by the common ratio -6. Like with the previous problem, we divide any term over its previous one to get the common ratio, so for example, term2/term1 = -12/2 = -6 = common ratio.

3 0
3 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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