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2 years ago

The expression 3(2x + 7) − 4(−3x + 5) is equal (=) to

Mathematics

2 answers:

Westkost [7]2 years ago

8 0

Answer:

-6x+26

Step-by-step explanation:

6x+21-12x+5

-6x+26

Anni [7]2 years ago

4 0

Answer:

3(2x+7) - 4 (-3x+5) = 18x+1

Step-by-step explanation:

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7 times z reduced by a third of the product

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the answer to 7 times z reduced by a third of the product is 2z

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1 year ago

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Name the value of each 2 in 222,222?

Eva8 [605]

Hundred thousand,ten thousand,hundred,ten,ones

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1 year ago

Markup = original cost × percentage markup selling price = original cost + markup The manufacturing cost of a cell phone was $12

strojnjashka [21]

C.) Old Price: $148.80; New Prices: $142.60

Using The Formula Given

M = O x P = O + M

Markup Value After 20% Is $148.80

Markup Value After 15% is $142.60

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1 year ago

Determine if each of the following sets is a subspace of ℙn, for an appropriate value of n. Type "yes" or "no" for each answer.

xxMikexx [17]

Answer:

1. Yes.

2. No.

3. Yes.

Step-by-step explanation:

Consider the following subsets of Pn given by

1.Let W1 be the set of all polynomials of the form $p(t)=at^2$ , where a is in ℝ.

2.Let W2 be the set of all polynomials of the form $p(t)=t^2+a$ , where a is in ℝ.

3. Let W3 be the set of all polynomials of the form $p(t)=at^2+at$ , where a is in ℝ.

Recall that given a vector space V, a subset W of V is a subspace if the following criteria hold:

- The 0 vector of V is in W.

- Given v,w in W then v+w is in W.

- Given v in W and a a real number, then av is in W.

So, for us to check if the three subsets are a subset of Pn, we must check the three criteria.

- First property:

Note that for W2, for any value of a, the polynomial we get is not the zero polynomial. Hence the first criteria is not met. Then, W2 is not a subspace of Pn.

For W1 and W3, note that if a= 0, then we have p(t) =0, so the zero polynomial is in W1 and W3.

- Second property:

W1. Consider two elements in W1, say, consider a,b different non-zero real numbers and consider the polynomials

$p_1 (t) = at^2, p_2(t)=bt^2$ .

We must check that p_1+p_2(t) is in W1.

Note that

$p_1(t)+p_2(t) = at^2+bt^2 = (a+b)t^2$

Since a+b is another real number, we have that p1(t)+p2(t) is in W1.

W3. Consider two elements in W3. Say $p_1(t) = a(t^2+t), p_2(t)= b(t^2+t)$ . Then

$p_1(t) + p_2(t) = a(t^2+t) + b(t^2+t) = (a+b) (t^2+t)$

So, again, p1(t)+p2(t) is in W3.

- Third property.

W1. Consider an element in W1 $p(t) = at^2$ and a real scalar b. Then

$bp(t) = b(at^2) = (ba)t^2)$ .

Since (ba) is another real scalar, we have that bp(t) is in W1.

W3. Consider an element in W3 $p(t) = a(t^2+t)$ and a real scalar b. Then

$bp(t) = b(a(t^2+t)) = (ba)(t^2+t)$ .

Since (ba) is another real scalar, we have that bp(t) is in W3.

After all,

W1 and W3 are subspaces of Pn for n= 2

and W2 is not a subspace of Pn.

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2 years ago

High school students across the nation compete in a financial capability challenge each year by taking a National Financial Capa

Misha Larkins [42]

Answer:

<u>The correct answer is that a student have to score 1.41 standard deviations above the mean to be publicly recognized.</u>

Step-by-step explanation:

For answering the question, we don't know the score mean of the National Financial Capability Challenge Exam and neither the population or number of students who take the exam. The only information provided is that the public recognition in this normal distribution is only for students that scores in the top 8%. In other words for students above 92% of the population.

With this information, we can go to a Z Score Table and check that for being on the top 8% (above the 92% of any population), your result must be 1.405 standard deviations above the mean.

<u>Rounding the answer to 2 decimal places, it's 1.41 standard deviations above the mean.</u>

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2 years ago