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3 years ago

The polynomial 8x2 – 8x + 2 – 5 + x is simplified to 8x2 – gx – h. What are the value of g and h?

Mathematics

1 answer:

igor_vitrenko [27]3 years ago

3 0

Answer:

Option D: g = 7 and h = 3

Step-by-step explanation:

The polynomial is;

8x² – 8x + 2 – 5 + x

Simplifying this gives;

8x² - 7x - 3

If this is simplified to 8x² – gx – h

Thus, by comparison of terms;

– gx = - 7x

-x will cancel out to get;

g = 7

Similarly, - h = - 3

Thus,h = 3

Therefore; g = 7 and h = 3

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Answer:

a. m<B = 100°

b. m<L = 55°

c. Scale factor = 9/11

Step-by-step explanation:

a) Similar triangles have their three corresponding angles congruent to each other, while the ratio of their corresponding sides are proportional to each other.

Since ∆ABC ~ ∆GLJ, therefore,

<A = <G,

<B = <J

<C = <L

m<J = 100° (given)

Therefore,

m<B = m<J = 100°

m<B = 100°

b) m<A = m<G

m<A = 25° (given)

Therefore,

m<A = m<G = 25°

m<G = 25°

m<L = 180° - (m<J + m<G)

Substitute

m<L = 180° - (100° + 25°)

m<L = 55°

c) scale factor of smaller triangle to the larger = side length of smaller triangle / corresponding side length of bigger triangle

Scale factor = AB/GJ

Substitute

Scale factor = 18/22

Simplify

Scale factor = 9/11

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Why is 7.8 an irrational number?

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Slope = 1.75; (10, 16.5)

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2 years ago

Find all possible values of α+

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Answer:

$\rm\displaystyle 0,\pm\pi$

Step-by-step explanation:

please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation

===========================

we want to find all possible values of α+β+γ when tanα+tanβ+tanγ = tanαtanβtanγ to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma )$

factor out tanγ:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)$

divide both sides by tanαtanβ-1 and that yields:

$\rm\displaystyle \tan( \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) - 1}$

multiply both numerator and denominator by-1 which yields:

$\rm\displaystyle \tan( \gamma ) = - \bigg(\frac{ \tan( \alpha ) + \tan( \beta ) }{ 1 - \tan( \alpha ) \tan( \beta ) } \bigg)$

recall angle sum indentity of tan:

$\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta )$

let α+β be t and transform:

$\rm\displaystyle \tan( \gamma ) = - \tan( t)$

remember that tan(t)=tan(t±kπ) so

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi )$

therefore when k is 1 we obtain:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi )$

remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi )$

recall that if we have common trigonometric function in both sides then the angle must equal which yields:

$\rm\displaystyle \gamma = - \alpha - \beta \pm \pi$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }$

when is 0:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 )$

likewise by Opposite Angle Identity we obtain:

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 )$

recall that if we have common trigonometric function in both sides then the angle must equal therefore:

$\rm\displaystyle \gamma = - \alpha - \beta \pm 0$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ 0 }$

and we're done!

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