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noname [10]
3 years ago
6

Please help with multiple choice question

Mathematics
1 answer:
Volgvan3 years ago
7 0

Answer:

D:   {9, 21, 33}

Step-by-step explanation:

Ken worked 2, 8 and 14 hours on 3 separate days.

For working 2 hours, his earnings were f(2) = 2(2) + 5, or 9;

For working 8 hours, his earnings were f(28) = 2(8) + 5, or 21; and

For working 14 hours, his earnings were f(14) = 2(14) + 5, or 33

Thus, the range of this function for the days given is {9, 21, 33} (Answer D)

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Enter a number in each blank M(x)=(2x-6)(x-4) true statements when M(x)=0 when x=
KengaRu [80]

Answer:

{3, 4}

Step-by-step explanation:

"M(x)=(2x-6)(x-4) true statements when M(x)=0 when x= ?" asks us to find the "roots" of M(x); that is, the x values at which M(x) = 0.  Thus, we set

(2x - 6)(x - 4) = 0, which is equivalent to 2(x - 3)(x - 4) = 0.

Thus, x - 3 = and x = 3; also x - 4 = 0, so that x = 4.  

The roots of M(x) are {3, 4}

Using the language of the original problem:  "true statements when M(x)=0 when x="              the correct results, inserted into the blanks, are x = 3 and x = 4.

4 0
3 years ago
Find the roots of the polynomial equation. x^2+2x−48=0
Zanzabum
(x+8)(x-6)= 0
x^2-48+8x-6x=0
x^2-48+2x=0
x+8=0 x-6=0
x= -8 x=6
The answer is -8 and 6, I did the process in case you need it
3 0
3 years ago
I need help with geometry
andrew-mc [135]
What do you need help with?
8 0
4 years ago
Read 2 more answers
Which number below is a perfect square? 100 90 50 75 ​
Umnica [9.8K]
100








ggbnjj mj ddcnmikiyewazxbnoorr
4 0
3 years ago
I NEED HELP ON THIS PROBLEM!!! PLEASE
Allushta [10]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

5x+8y=-9\implies 8y=-5x-9\implies y=\cfrac{-5x-9}{8} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{5}{8}} x-9\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}

well then, so since this equation has that slope therefore

\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{-5}{8}} ~\hfill \stackrel{reciprocal}{\cfrac{8}{-5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{8}{-5}\implies \cfrac{8}{5}}}

so we're really looking for the equation of a line whose slope is 8/5 and runs through (10,10)

(\stackrel{x_1}{10}~,~\stackrel{y_1}{10}) ~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{8}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{\cfrac{8}{5}}(x-\stackrel{x_1}{10})

5 0
3 years ago
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