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Sidana [21]
3 years ago
6

Show all work to multiply (2+√-25)(4-√-100)

Mathematics
1 answer:
ki77a [65]3 years ago
7 0

Answer:

<h3>-42</h3>

Step-by-step explanation:

given the expression;

(2+√-25)(4-√-100)

Note that √-25 = √-1 * √25

√-100 = √-1 * √100

According to complex number,  √-1 = i

√-25 = 5i

√-100 = 10i

The expression becomes

(2+5i)(4-10i)

Open the parentehsis;

= 2(4)-20i+20i-50i²

=  8-50i²

Since i² = -1

The expression becomes;

8-50i² = 8-50(-1)²

8-50i² = 8-50 = -42

Hence the product of the expression is -42

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Answer:

the MEAN for class a is 72.6

the MEAN for class b is 84.0

if you take these means and compare them, you can see that on average, class B performed better.

the MEDIAN for class a is 72

the MEDIAN for class b is 85

Step-by-step explanation:

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2 years ago
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Answer:

\frac{1}{5}

Step-by-step explanation:

Using the rules of exponents

a^{m} × a^{n} = a^{(m+n)}, \frac{a^{m} }{a^{n} } = a^{(m-n)}, (a^m)^{n} = a^{mn}

Simplifying the product of the first 2 terms

\frac{a^{p^2+pq} }{a^{pq+q^2} } × \frac{a^{q^2+qr} }{a^{qr+r^2} }

= a^{p^2-q^2} × a^{q^2-r^2}

= a^{p^2-r^2}

Simplifying the third term

5((a^p+r)^{p-r}

= 5a^{(p+r)(p-r)} = 5a^{(p^2-r^2)}

Performing the division, that is

\frac{a^{(p^2-r^2)} }{5a^{(p^2-r^2)} } ← cancel a^{(p^2-r^2)} on numerator/ denominator leaves

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A coin is tossed 6 times how many outcomes are possible
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emmainna [20.7K]

Given:

The scientific notation of a number is 1.3\times 10^5.

To find:

The standard form of the given number.

Solution:

Standard form: The normal way of writing a number is called standard form. For example: 200, 25600, 1025000 etc.

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1.3\times 10^5

It can be written as

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