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-Dominant- [34]
3 years ago
11

HELP ME PLEASSSSSSSSSSEEEEEEEEEEEEEEEEEEEE

Mathematics
1 answer:
Contact [7]3 years ago
7 0

Answer:

105

Step-by-step explanation:

because you multiply 5 times 3 which is 15 times 7 equals 105

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492 divided by 41= 41. He can make 41 full bouquets.
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How to evaluate (9/16)^-2
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The correct answer is b. When raising a number to a negative power switches the denominator and the numerator.
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A model of a skyscraper is 1.6 inches long, 2.8 inches wide, and 11.2 inches high. The scale factor is 8 inches = 250 feet. What
kozerog [31]

Answer:

50 feet long

87.5 feet wide

350 feet high

Step-by-step explanation:

8 inches : 250 feet

1 inch : 250/8 feet

250/8 = 31.25

Length: 31.25 × 1.6 = 50 feet

Width: 31.25 × 2.8 = 87.5 feet

Height: 31.25 × 11.2 = 350 feet

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N76 [4]

Answer:

4 photos on each page

Step-by-step explanation:

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Determine if the described set is a subspace. Assume a, b, and c are real numbers. The subset of R3 consisting of vectors of the
Arte-miy333 [17]

This question is incomplete, the complete question is;

Determine if the described set is a subspace. Assume a, b, and c are real numbers.

The subset of R³ consisting of vectors of the form \left[\begin{array}{ccc}a\\b\\c\end{array}\right] , where abc = 0

  • The set is a subspace
  • The set is not a subspace

Answer:

Therefore; The set is not a subspace

Step-by-step explanation:

Given the data the question;

the subset R³;

S = {  \left[\begin{array}{ccc}a\\b\\c\end{array}\right] , where abc = 0 }

we know that a subset of R³ is a subspace if it stratifies the following properties;

  1. it contains additive identity
  2. it is closed under addition
  3. it is closed under scales multiplication

Looking at the properties, we can say that it is not a subspace

As;

         u = \left[\begin{array}{ccc}1\\1\\0\end{array}\right] ∈ S       and v = \left[\begin{array}{ccc}0\\1\\1\end{array}\right]  ∈ S

As    1×1×0=0                          0×1×1=0

But   u+v = \left[\begin{array}{ccc}1+0\\1+1\\0+1\end{array}\right] = \left[\begin{array}{ccc}1\\2\\1\end{array}\right]  ∉  S     as 1×2×1 ≠ 0

Hence, it is not closed under addition.

Therefore; The set is not a subspace

4 0
3 years ago
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