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3 years ago

Kian has been asked to create a company logo to be put on all company merchandise. The logo needs to be

Mathematics

1 answer:

Sladkaya [172]3 years ago

3 0

Kian has to work with 6 square inches

Step-by-step explanation:

Given that the logo has to be created in a rectangular shape

Length = 3 inches

Width = 2 inches

We have to calculate the area of the given space to find out that how much area Kian will work with.

So,

$Area = Length*Width\\=2*3\\=6\ in^2$

Hence,

Kian has to work with 6 square inches

Keywords: Rectangle, Area

Learn more about rectangle at:

#LearnwithBrainly

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erica [24]

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Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,

posledela

Answer:

number of multiplies is n!
n=10, 3.6 ms
n=15, 21.8 min
n=20, 77.09 yr
n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

mpy[n] = n·mp[n-1]

mpy[2] = 2

So, ...

mpy[n] = n! . . . n ≥ 2

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

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For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

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3 years ago

In VWX x=41 cm v=17 cm and W=134 degrees. Find the length of w, the the nearest centimeter

pav-90 [236]

Answer:

Step-by-step explanation:

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3 years ago

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3 years ago

The sum of the first 150 negative integers is represented using the expression What is the sum of the first 150 negative integer

Kamila [148]

Answer:

C. -11,325

Step-by-step explanation:

To know the answer, is convenient to replace some values of "n" in the sum $\sum_{(n=1)}^{150}[-1-(n-1)]$ .

The result would appear after adding up every value of the expression $\sum_{(n=1)}^{150}[-1-(n-1)]$ when n=1,2,3.....,150.
When n=1, the expression takes the value of (-1): $[-1-(1-1)]= (-1)-0=-1$ .
When n=2, the expression takes the value of (-2): $[-1-(2-1)]=-1-1=-2$ .
Following this way, for every n, we will obtain -n, then, the sum will be: $-1-2-3-4-5-6-...-150$ . This sum can actually be expressed as $\sum_{n=1}^{150}(-n)$ , which is the result of solving the initial expression of the sum $-1-(n-1)=-1-n+1=-n$ .
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