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MaRussiya [10]
3 years ago
14

What is the missing length+ Work?

Mathematics
1 answer:
Deffense [45]3 years ago
8 0
Answer - height                                                                                                                                                
at least thats what i think
You might be interested in
(2x³+4x²-x+8) / (x² + 3x + 2)
a_sh-v [17]

\dfrac{2x^3+4x^2 -x+8}{x^2 +3x+2}\\\\\\=\dfrac{2x^3+6x^2+4x-2x^2-6x-4+x+12}{}\\\\=\dfrac{2x(x^2 +3x+2)-2(x^2 +3x+2)+x+12}{x^2 +3x+2}\\\\\\=\dfrac{2x(x^2+3x+2)}{x^2 +3x +2 }- \dfrac{2(x^2 +3x +2)}{x^2 +3x +2 } + \dfrac{x+12}{x^2 +3x+2}\\\\\\=2x-2 + \dfrac{x+2}{x^2 +3x +2}\\\\\\=2x-2 + \dfrac{x+12}{x^2 +2x +x +2}\\\\\\=2x-2 + \dfrac{x+12}{x(x+2) + x+2}\\\\\\=2x-2 + \dfrac{x+12}{(x+1)(x+2)}\\

3 0
1 year ago
Let Z={a,c,{a,b}}. What is |Z|?
Lina20 [59]

Z=\{a,c,\{a,b\}\}

\boxed{|Z|=3} (treat \{a,b\} as one element of Z)

The power set of Z is

\boxed{2^Z=\bigg\{\{\},\{a\},\{c\},\big\{\{a,b\}\big\},\{a,c\},\big\{a,\{a,b\}\big\},\big\{c,\{a,b\}\big\},\big\{a,c,\{a,b\}\big\}\bigg\}}

1. \{a,c\}\subseteq Z is true because both a\in Z and c\in Z.

2. a\in Z is true.

3. \{c\}\subseteq Z is true (same reason as part 1).

4. \{c\}\in Z is false because Z does not contain the set \{c\}, rather just the element c itself.

5. b\in Z is false because the element b on its own simply is not in Z. That b\in\{a,b\} does not mean b\in Z, but that b belongs to a subset of Z.

6. \{a,b\}\in Z is true.

6 0
3 years ago
If the common difference of an ap is 3/2 and its 20 th term 35×1/2 find first term and 15 th term
Lady bird [3.3K]

Answer:

Step-by-step explanation:

d = 3/2

a₂₀ = a₁+19d

35/2 = a₁ + 19×3/2

a₁ = 35/2 - 19×3/2 = -11

a₁₅ = a₁+14d = -11 + 14×3/2 = 10

4 0
2 years ago
Which expression represents the number 2i4−5i3+3i2+−81‾‾‾‾√ rewritten in a+bi form?
vichka [17]

Answer:

The expression -1+14i represents  the number 2i^4-5i^3+3i^2+\sqrt{-81} rewritten in a+bi form.

Step-by-step explanation:

The value of i is i=\sqrt{-1}[tex] or [tex]i^{2}=-1[\tex].Now [tex]i^{4} in term of i^{2}[\tex] can be written as, [tex]i^{4}=i^{2}\times i^{2}

Substituting the value,

i^{4}=\left(-1\right)\times \left(-1\right)

Product of two negative numbers is always positive.

\therefore i^{4}=1

Now i^{3} in term of i^{2}[\tex] can be written as, [tex]i^{3}=i^{2}\times i

Substituting the value,

i^{3}=\left(-1\right)\times i

Product of one negative  and one positive numbers is always negative.

\therefore i^{3}=-i

Now \sqrt{-81} can be written as follows,

\sqrt{-81}=\sqrt{\left(81\right)\times\left(-1\right)}

Applying radical multiplication rule,

\sqrt{ab}={\sqrt{a}}\sqrt{b}

\sqrt{\left(81\right)\times\left(-1\right)}={\sqrt{81}}\sqrt{-1}

Now, \sqrt{\left(81\right)=9 and \sqrt{-1}}=i

\therefore \sqrt{\left(81\right)\times\left(-1\right)}=9i

Now substituting the above values in given expression,

2i^4-5i^3+3i^2+\sqrt{-81}=2\left(1\right)-5\left(-i\right)+3\left(-1\right)+9i

Simplifying,

2+5i-3+9i

Collecting similar terms,

2-3+5i+9i

Combining similar terms,

-1+14i

The above expression is in the form of a+bi which is the required expression.

Hence, option number 4 is correct.

5 0
3 years ago
Please help and thank you
Black_prince [1.1K]

I'm honestly not really sure, but it has to be either (2,10) or (10,2). I've never seen a question like this...

Answer:

Step-by-step explanation:

8 0
3 years ago
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