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MariettaO [177]
3 years ago
11

Math hw graded on accuracy pls help!!

Mathematics
1 answer:
lubasha [3.4K]3 years ago
8 0

Answer:

  see below

Step-by-step explanation:

Putting n=1 into the formulas eliminates the first two choices, since they evaluate to 3.57 and 0.85, respectively, not 4.2.

Putting n=2 into the formula eliminates the third choice, because you know that 4.2(0.15) < 3.57 without even using a calculator.

The only viable choice (and the correct choice) is the last one.

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Step-by-step explanation: 471,464

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Which statements about the local maximums and minimums for the given function are true? Choose three options.
Nostrana [21]

Answer:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.

Step-by-step explanation:

The true statements are:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.  

Lets discuss each option one by one:

Over the interval [1, 3], the local minimum is 0

This is a false statement. Look at the graph. The minimum point given is (3.4,-8). Therefore the local minimum is -8 not 0

Over the interval [2, 4], the local minimum is –8.

This statement is true because the given minimum point is(3.4, -8). Thus the  local minimum is -8 which is true

Over the interval [3, 5], the local minimum is –8.

According to the given minimum point, the local minimum  is -8 which is true

Over the interval [1, 4], the local maximum is 0.

Look at the graph. The maximum point given is (2,0). Thus this statement is true because local maximum is 0.

Over the interval [3, 5], the local maximum is 0.

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3 years ago
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Which coefficient matrix represents a system of linear equations that has a unique solution ?
finlep [7]

Answer:

Option C

Step-by-step explanation:

We are given a coefficient matrix along and not the solution matrix

Since solution matrix is not given we cannot check for infinity solutions.

But we can check whether coefficient matrix is 0 or not

If coefficient matrix is zero, the system is inconsistent and hence no solution.

Option A)

|A|=\left[\begin{array}{ccc}4&2&6\\2&1&3\\-2&3&-4\end{array}\right] =0

since II row is a multiple of I row

Hence no solution or infinite

OPtion B

|B|=\left[\begin{array}{ccc}2&0&-2\\-7&1&5\\4&-2&0\end{array}\right] \\=2(10)-2(10)=0

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Option C

\left[\begin{array}{ccc}6&0&-2\\-2&0&6\\1&-2&0\end{array}\right] \\=2(36-2)=68

Hence there will be a unique solution

Option D

\left[\begin{array}{ccc}5&10&5\\4&1&4\\-1&-2&-1\end{array}\right] \\=2(10)-2(10)=0=0

(since I row is -5 times III row)

Hence there will be no or infinite solution

Option C is the correct answer



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3 years ago
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