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Kobotan [32]
4 years ago
9

Determine whether the following function is linear or quadratic. Identify the quadratic, linear, and constant terms. f(x)=3x(x−1

)−(3x+7)
Mathematics
1 answer:
Nataliya [291]4 years ago
6 0
It is quadratic, because the highest degree for x is 2, x². Expand the expression and you'll see: f(x)=3x²-3x-3x-7
simplify: f(x)=3x²-6x-7
the quadratic term is 3x², the linear term is -6x, the constant is -7
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Please help me :(bbbbbb
Strike441 [17]

Answer:

im sorry this seems to not be correctly stated

Step-by-step explanation:

4 0
3 years ago
-1/19,7/4,-4/7 least to greatest
oee [108]
Answer:

Order from Least to Greatest
-4/7  <  -1/19  <  7/4


Showing Work
Rewriting as fractions or any negatives if necessary:
-1/19, 7/4, -4/7

The least common denominator (LCD) is: 532.

Rewriting as equivalent fractions with the LCD:
-28/532, 931/532, -304/532

Ordering these fractions by the numerator:
-304/532  <  -28/532  <  931/532

Therefore, the order of your input is:
-4/7  <  -1/19  <  7/4

8 0
3 years ago
Determine algebraically whether the function is even, odd, or neither even nor odd.
ololo11 [35]

Answer:

odd

Step-by-step explanation:

Just so you know there are shortcuts for determining if a polynomial function is even or odd. You just to make sure you use that x=x^1 and if you have a constant, write it as constant*x^0 (since x^0=1)

THEN!

If all of your exponents are odd then the function is odd

If all of your exponents are even then the function is even

Now you have -4x^3+4x^1

3 and 1 are odd it is an odd function

This a short cut not the legit algebra way

let me show you that now:

For it to be even you have f(-x)=f(x)

For it be odd you have f(-x)=-f(x)

If you don't have either of those cases you say it is neither

So let's check

plug in -x  -4(-x)^3+4(-x)=-4*-x^3+-4x=-4x^3+-4x

that's not the same so not even

with if we factor out -1 .... well if we do that we get -(4x^3+4x)=-f(x)

so it is odd.

6 0
4 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}\\\\&#10;slope=\cfrac{a}{{{ b}}}\qquad negative\implies  -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\&#10;-------------------------------\\\\

\bf \boxed{5i+12j}\implies &#10;\begin{array}{rllll}&#10;\ \textless \ 5&,&12\ \textgreater \ \\&#10;x&&y&#10;\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}&#10;\\\\\\&#10;slope=\cfrac{12}{{{ 5}}}\qquad negative\implies  -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}&#10;\\\\\\&#10;\ \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}}&#10;\\\\\\&#10;\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}&#10;\\\\\\&#10;\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}
4 0
4 years ago
Abigail went shopping for her sister's birthday. She spent 2/7 of her money on a necklace and 3/4 of the remainder on new outfit
DochEvi [55]
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4 0
3 years ago
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