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3 years ago

Is (x - 1) a factor of P(x) = 4x? - 9x2 + 4x+1? If it is, write P(x) as a product of

Mathematics

1 answer:

trasher [3.6K]3 years ago

8 0

Answer: i don't even know what a factor is

Step-by-step explanation:

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3p + 2q = 14<br> 10p + 6q = 44<br> What is p and what is q

Airida [17]

Answer:

p = 2 ; q = 4

Step-by-step explanation:

Given tbe equation :

3p + 2q = 14 - - - (1)

10p + 6q = 44 - - -(2)

What is p and what is q

This is a simultaneous equation ; using elimination method :

Multiply (1) by 6 and (2) by 2

18p + 12q = 84 - - - - (3)

20p + 12q = 88 - - - (4)

Subtract (3) and (4)

-2p = - 4

p = 4/2

p = 2

Put p = 2 in (1)

3p + 2q = 14

3(2) + 2q = 14

6 + 2q = 14

2q = 14 - 6

2q = 8

q = 8/2

q = 4

p = 2 ; q = 4

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3 years ago

HELP ASAP! GIVING BRAINLIEST

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Step-by-step explanation: The Answer Would Be D

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A candy factory made 54 boxes of chocolates. Each box weighs 2 pounds. They packed the boxes in 6 cases with the same number of

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There was 18 pounds of chocolate in each case

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3 years ago

How do you find a vector that is orthogonal to 5i + 12j ?

Rashid [163]

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
-------------------------------\\\\$

$\bf \boxed{5i+12j}\implies 
\begin{array}{rllll}
\ \textless \ 5&,&12\ \textgreater \ \\
x&&y
\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}
\\\\\\
slope=\cfrac{12}{{{ 5}}}\qquad negative\implies -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}
\\\\\\
\ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}$

if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

$\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}
\\\\\\
\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}
\\\\\\
\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}$

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3 years ago