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Lorico [155]
4 years ago
12

What is the inverse of f(x)=6x^2

Mathematics
1 answer:
likoan [24]4 years ago
8 0

Answer:

f^-1 (x) = √ 6x/6 - √ 6x/6

Step-by-step explanation:

interchange the variables and solve for y.

sorry its a little hard to read i didnt have all the symbols on my keyboard

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a bicyclist covered 5/7 of his route and an additional 40 miles. he has yet to cover 118 miles less than 0.75 of his route. how
Oduvanchick [21]

The length of the route of a bicylist that has covered 5/7 of his route and an additional 40 miles and yet to cover 118 miles less than 0.75 of his route is 168 miles.

<h3>How to form equation and solve for the variable?</h3>

The bicyclist covered 5 / 7 of his route and an addditional 40 miles.

Let

x = distance of his route in miles

Hence,

distance covered = 5 / 7 x + 40

He has yet to cover 118 miles less than 0.75 of his route. Therefore,

distance not covered = 0.75x - 118

Distance of his route = 5 / 7 x + 40 + 0.75x - 118

x = 5 / 7 x + 40 + 0.75x - 118

x = 1.464x - 78

x -  1.464x = -78

-0.464x = -78

x = 78 / 0.464

x = 168.103448276

x = 168 miles

learn more on equation here: brainly.com/question/13887440

8 0
3 years ago
Which statements can be used to justify the fact that two right angles are supplementary? Check all that apply.
mamaluj [8]
The answer is a and c
5 0
3 years ago
Read 2 more answers
BRAINLIEST AND 100 POINTS. Please help by explaining how you got the answer
spayn [35]

Answer:

it may be A log x ........

4 0
3 years ago
The friendly sausage factory (fsf) can produce hot dogs at a rate of 5,000 per day. fsf supplies hot dogs to local restaurants a
juin [17]

Answer:

a. 21 327 hot dogs/run

b. 70 runs/yr

c. 4 da/run

Step-by-step explanation:

Data:

Production rate (p)           = 5000/da

Usage rate (u)                  =    260/da

Setup cost (S)                   = $66

Annual carrying cost (H) = $0.45/hot dog

Production days (d)         = 294 da

Calculations:

a. Optimal run size

(i) Annual demand (D) = pd = (5000 hot dogs/1 day) × (294 days/1 yr)

= 1 470 000 hot dogs/yr

(ii) Economic run size

Q_{0}= \sqrt{\frac{2DS }{ h}\times\frac{ p}{p-u }}

= \sqrt{\frac{2\times1470000\times66 }{ 0.45}\times\frac{ 5000}{5000-260 }}

= \sqrt{431200000\times\frac{ 5000}{4740 }}

= \sqrt{454852321}

= 21 327 hot dogs/run

b. Number of runs per year

Runs = D/Q₀ = (1 470 000 hot dogs/1yr) × (1 run/21 327 hotdogs)

= 70 runs/yr

c. Length of a run

Length = Q₀/p = (21 327 hot dogs/1 run) × (1 da/5000 hot dogs)

= 4 da/run

8 0
3 years ago
Consider the line
AveGali [126]
Y - y₁ = m(x - x₁)
y - 5 = ⁵/₆[x - (-6)]
y - 5 = ⁵/₆(x + 6)
y - 5 = ⁵/₆(x) + ⁵/₆(6)
y - 5 = ⁵/₆x + 5
  + 5          + 5
     y = ⁵/₆x + 10

y - y₁ = m(x - x₁)
y - 5 = -1¹/₅[x - (-6)]
y - 5 = 1¹/₅(x + 6)
y - 5 = 1¹/₅(x) + 1¹/₅(6)
y - 5 = 1¹/₅x + 7¹/₅
  + 5           + 5
     y = 1¹/₅x + 12¹/₅
8 0
3 years ago
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