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2 years ago

What do I do with the “y = -f(x) +6” given to find the Domain and Range?

Mathematics

2 answers:

Bond [772]2 years ago

7 0

Answer: hvfiikncdnf;oewh;d..ew

Step-by-step explanation:

Kamila [148]2 years ago

5 0

Answer:

Step-by-step explanation:

a.Changing f with -f does not change its domain, it just flips the image across the x axis. Adding 6 to the result, just shifts everything towards the positive y by 6 units. Domains stays the same, from -4 to +8. Range instead goes up, the lowest value is -2+6 = 4 and the highest value is 2+6 = 8.

So domain remains the same [-4; 8] and range goes to [4;8]

a. This time we're shifting everything on the horizontal axis, one unit to the left (the way to remember it is that f(x) has a minimum at x=4, while after the shift the new value happens at x=3) multiplying the value by 3 means that every value gets scaled up. It means the minimum becomes -2*3= -6 and the maximum happens at 2*3= 6. The shift brings the domain to [-5;7] and the range to [-6,6]

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LMN is a right-angle triangle. Angle NLM=90. PQ is parallel to LM. The area of triangle PNQ is 8cm^2. The area of triangle LPQ i

likoan [24]

Answer:

The area of LQM is $48cm^2$

Step-by-step explanation:

Given

Area of PNQ = 8

Area of LPQ = 16

See attachment for triangles

The area of PNQ is calculated as:

$Area = \frac{1}{2} * PQ * PN$

Substitute 8 for Area

$8 = \frac{1}{2} * PQ * PN$

$PQ * PN = 16$

The area of LPQ is calculated as:

$Area = \frac{1}{2} * PQ * LP$

Substitute 16 for Area

$16= \frac{1}{2} * PQ * LP$

From the attachment:

$PN + LP =LN$

Make LP the subject

$LP = LN -PN$

So:

$16= \frac{1}{2} * PQ * (LN -PN)$

We have:

$16= \frac{1}{2} * PQ * (LN -PN)$ and $PQ * PN = 16$

Equate both expressions:

$\frac{1}{2} * PQ *(LN - PN) = PQ * PN$

Divide both sides by PQ

$\frac{1}{2} (LN - PN) = PN$

Multiply both sides by 2

$LN - PN = 2PN$

$LN= 3PN$

Since PNQ is similar to LNM, the following equivalent ratios exist:

$\frac{LM}{PQ} = \frac{LN}{PN}$

Substitute $LN= 3PN$

$\frac{LM}{PQ} = \frac{3PN}{PN}$

$\frac{LM}{PQ} = 3$

$LM = 3PQ$

Area of LQM is:

$Area = \frac{1}{2} * LM * LP$

This gives:

$Area = \frac{1}{2} * 3PQ * LP$

$Area = 3 *\frac{1}{2} *PQ * LP$

Recall that:

$16= \frac{1}{2} * PQ * LP$

So:

$Area = 3 *16$

$Area = 48cm^2$

5 0

3 years ago

Can you answer question 45 and 46? (Show work)

muminat

45. These word problems often seem to be two equations in two unknowns, here h for hits and m for misses.

"After 30 tries" translates into

$h + m = 30$

+5 for a hit, -10 for a miss, John lost 90 translates to

$5h + -10 m = -90$

We have

$m = 30 - h$

$5h - 10(30 -h) = -90$

$5h - 300 + 10h = -90$

$15h = 210$

$h = 210/15 = 14$

Answer: 14

Check: 5(14)-10(16)=-90, good

46. We have to make the odd assumption the grocer sells for his cost, but let's not beat up the question too hard.

Again, two unknowns, b kilograms of Brazil nuts and c kilograms of cashews.

... sold 12 kg of the mixture ...

$b + c = 12$

... cashews $8/kg and Brazil nuts $10/kg and the results is $8.50/kg ...

This one's a bit more involved to translate because we have to recognize the total price is the 8.50 per kg times 12 kg.

$8c + 10b = 12(8.50) = 102.00$

The solving is pretty much just like the last one; we eliminate the unknown we're not asked for.

$b = 12 - c$

$8c + 10(12 - c) = 102$

$-2c = 102 - 120 = -18$

$c = 9$

Answer: 9

Check: (9(8)+10(3))/12=102/12 = 8.50 good

4 0

3 years ago

When are x2+2x/5x and it's simplified form equal

klio [65]

Answer:

The correct answer is

${x}^{2} + \frac{2}{5}$

3 0

2 years ago

If the basalt at the top of the ancient valley below Vulcan's Throne yields a radiometric date (K-Ar) of 15,000 years, and the l

zepelin [54]

Answer:

$1.69\times 10^3$ feet/million year

Step-by-step explanation:

We are given that

Elevation of top of basalt in valley =4100 feet

Elevation of bottom of basalt in valley =2100 feet

Age of top of basalt in valley=15000 years

Age of bottom of basalt in valley=1.2 million years= $1.2\times 10^6 years=1200000 years$

We have to find the rate of basalt filling in the valley in feet per million yeas

Rate of basalt filling in the valley= $\frac{Elevation\;of\;top\;of\;basalt\;in\;valley-elevation\;of\;bottom\;of\;basalt\;in\;valley}{age\;of\;bottom\;of\;basalt\;in\;valley-age\;of\;top\;of\;basalt\;in\;valley}$

Rate of basalt filling in the valley= $\frac{4100-2100}{1200000-15000}=\frac{2000}{1185000}=1.69\times 10^{-3} feet/year$

Rate of basalt filling in the valley= $1.69\times 10^{-3}\times 10^{6}=1.69\times 10^3$ feet/million year

6 0

3 years ago

ANSWER QUICK! MANY POINTS! NO WRONG ANSWERS PLEASE!

g100num [7]

Answer:

see below

Step-by-step explanation:

Function 1

Find the slope

m = ( y2-y1)/ (x2-x1)

= (-5 -3) /(3 - -1)

= ( -5-3)/(3+1)

= -8/4

= -2

Function 2

Using the points (0,4) and (-1,0)

m = ( y2-y1)/ (x2-x1)

m = ( 0-4)/(-1-0)

= -4/-1

= 4

The function with the greater rate of change is function 2

3 0

2 years ago