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Natali [406]
2 years ago
6

I need help on this problem! g(h+2/3)=1 for h

Mathematics
1 answer:
Tju [1.3M]2 years ago
7 0

Answer:

g(h +  \frac{2}{3} ) = 1 \\  \\ gh +  \frac{2}{3} g = 1 \\  \\ gh = 1 -  \frac{2}{3} g \\  \\  h = (1 -  \frac{2}{3} g )  \div g \\  \\ h = (1 -  \frac{2}{3} g ) \times   \frac{1}{g}  \\  \\ h =  \frac{1}{g}  -  \frac{2}{3}

I hope I helped you^_^

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Examples of properties communative, associative, and identity?
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What’s the answer?and how do you get it
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the solid is made up of 2 regular octagons, 8 sides, joined up by 8 rectangles, one on each side towards the other octagonal face.

from the figure, we can see that the apothem is 5 for the octagons, and since each side is 3 cm long, the perimeter of one octagon is 3*8 = 24.

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\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}ap~~ \begin{cases} a=apothem\\ p=perimeter\\[-0.5em] \hrulefill\\ a=5\\ p=24 \end{cases}\implies A=\cfrac{1}{2}(5)(24)\implies \stackrel{\textit{just for one octagon}}{A=60} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{two octagon's area}}{2(60)}~~+~~\stackrel{\textit{eight rectangle's area}}{8(3\cdot 8)}\implies 120+192\implies 312

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3 years ago
List all possible rational roots. Then use synthetic division to confirm which rational roots exist:
Kisachek [45]

Answer:

\boxed{(1) \, x = \, \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10; (2) \, x = -2}

Step-by-step explanation:

2x³+ 6x² - x - 10 = 0

(1) Possible roots

The Rational Roots Theorem states that, if a polynomial has any rational roots, they will have the form p/q, where p is a factor of the constant term  and q is a factor of the leading coefficient.

\text{Possible rational root} = \dfrac{ p }{ q } = \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}}

In your function, the constant term is -10 and the leading coefficient is 2, so

\text{Possible root} = \dfrac{\text{factor of 10}}{\text{factor of 2}}

Factors of 10 = ±1, ±2, ±5, ±10

Factors of 2 = ±1, ±2

\text{Possible roots are } \large \boxed{\mathbf{x = \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10}}

(2) Synthetic division

Rather than work through all 12 possibilities, I will do one that works.

\begin{array}{r|rrrr}-2 & 2 & 6 & -1 & -10\\& & -4& -4 & 10\\& 2 & 2& -5 & 0\\\end{array}

So, x = -2 is a root, and the quotient is 2x² + 2x - 5.

(3) Check for other rational roots

2x² + 2x - 5 = 0

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Since irrational roots come in pairs, the cubic equation has two real, irrational roots and one rational root at x = -2.

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