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frozen [14]
2 years ago
5

2/3n + 5 greater than 12

Mathematics
2 answers:
Tanzania [10]2 years ago
6 0
2/3n + 5 > 12
2/3n > 12 - 5
2/3n > 7
n > 7 * 3/2
n > 21/2 or 10 1/2
tamaranim1 [39]2 years ago
3 0
The answer would be yes
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What is the value of z in the equation 4(2z + 3) = 12?<br><br> −13<br> 12<br> 0<br> 12
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Answer:

0

Step-by-step explanation:

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8z = 0

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4 0
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Alice is playing checkers against a computer and has won 8 games out of the 12 she’s played so far
den301095 [7]

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3 years ago
What does the following expression mean?<br> 7!<br><br> A. 7+6+5+4+3+2+1<br> B. 7·6·5·4·3·2·1
frez [133]

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Step-by-step explanation:

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3 0
3 years ago
Simplify the expressions and find the value:
Dahasolnce [82]

Answer:

x(x−3)+2=x.x−3x+2 \\ =x2−3x+2 \\F or x=1 \\ putting \:  x=1  \: in  \: expression \\ x2−3x+2=(1)2−3(1)+2 \\ =1−3+2 \\ =3−3 \\ =0

Solve  \: 3y(2y−7)−3(y−4)−63 for y=−2 \\ First \:  simplify \:  it, \\ =6y2−21y−3y+12−63 \\ =6y2−24y−51 \\ Put \:  y=−2 \: =6(−2)2−24(−2)−51  \\  =6×4+48−51 \: 6×4+48−51 \\  24+48−51 \\ =21

3 0
2 years ago
Solve the following equation by completing the square. 3x^2-3x-5=13
mr Goodwill [35]

we'll start off by grouping some

\bf 3x^2-3x-5=13\implies (3x^2-3x)-5=13\implies 3(x^2-x)-5=13 \\\\\\ 3(x^2-x)=18\implies (x^2-x)=\cfrac{18}{3}\implies (x^2-x)=6\implies (x^2-x+~?^2)=6

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?

well, let's recall that a perfect square trinomial is

\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²

\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}

6 0
3 years ago
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