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3 years ago

F(x)=6x-4 what is f(x)whence=8

Mathematics

1 answer:

pashok25 [27]3 years ago

3 0

Answer:

Step-by-step explanation:

Put 8 where x is and do the arithmetic.

f(8) = 6·8 -4 = 48 -4

f(8) = 44

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Need help ASAP please and thank you

likoan [24]

Answer:

The answer is C

$\frac{4 \times - 4 - 4}{ - 4 - 1}$

$\frac{ - 16 - 4}{ - 5}$

$\frac{ - 20}{ - 5} = 4$

6 0

3 years ago

If a, b, c are in A.P. show that<br>a (b + c)/bc,b(c + a) /ca, c(a-b )/bc<br>are in A.P.<br>

vfiekz [6]

Answer:

Step-by-step explanation:

$\frac{a(b+c)}{bc} ,\frac{b(c+a)}{ca} ,\frac{c(a+b)}{ab} ~are~in~A.P.\\if~\frac{ab+ca}{bc} ,\frac{bc+ab}{ca} ,\frac{ca+bc}{ab} ~are~in~A.P.\\add~1~to~each~term\\if~\frac{ab+ca}{bc} +1,\frac{bc+ab}{ca} +1,\frac{ca+bc}{ab} +1~are~in~A.P.\\if~\frac{ab+ca+bc}{bc} ,\frac{bc+ab+ca}{ca} ,\frac{ca+bc+ab\\}{ab} ~are~in~A.P.\\\\divide~each~by~ab+bc+ca\\if~\frac{1}{bc} ,\frac{1}{ca} ,\frac{1}{ab} ~are ~in~A.P.\\if~\frac{a}{abc} ,\frac{b}{abc} ,\frac{c}{abc} ~are~in~A.P.\\if~a,b,c~are~in~A.P.\\which~is~true.$

3 0

3 years ago

Last week at the park Haeather ran 4 miles in 48 minutes when she runs her next race she Hope's to improve her time by 2 minutes

Korvikt [17]

Answer:

Haeather ran at a speed of 5 miles per hour.

Step-by-step explanation:

Given that last week at the park Haeather ran 4 miles in 48 minutes, to determine what was Haeather rate of speed in miles per hour the following mathematical calculation must be performed:

48/4 = 12

Thus, Haeather ran 1 mile every 12 minutes.

60/12 = 5

1 x 5 = 5

Therefore, Haeather ran at a speed of 5 miles per hour.

3 0

3 years ago

3 + 3(k + 3) = 6(k - 2) +9

ANTONII [103]

Step-by-step explanation:

3 + 3(k + 3) = 6(k - 2) + 9

3 + 3k + 9 = 6k - 12 + 9

12 + 12 - 9 = 6k - 3k

15 = 3k

k = 15/3

k = 5

7 0

3 years ago

How do you find the area of the shaded region

Amiraneli [1.4K]

well, first off, let's notice that we have a trapezoid with a rectangle inside it, so the rectangle is really "using up" area that the trapezoid already has.

now, if we just get the area of the trapezoid, and then the area of the rectangle alone, and then subtract that area of the rectangle, the rectangle will in effect be making a hole inside the trapezoid's area, and what's leftover, is the shaded section, that part the hole is not touching.

$\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h&=height\\ a,b&=\stackrel{parallel~sides}{bases} \\\cline{1-2} h&=8\\ a&=16\\ b&=8 \end{cases}\implies A=\cfrac{8(16+8)}{2}\implies A=96 \\\\\\ \stackrel{\textit{area of the rectangle}}{(5\cdot 3)\implies 15}~\hfill \stackrel{~\hfill \textit{shaded area}}{\stackrel{\textit{trapezoid}}{96}~~ - ~~\stackrel{\textit{rectangle}}{15}~~ = ~~81}$

3 0

3 years ago