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2 years ago

Given f(x) and g(x) = f(x) + k, use the graph to determine the value of k.

Mathematics

1 answer:

Evgesh-ka [11]2 years ago

7 0

Answer:

k = 3

Step-by-step explanation:

The value of k is the amount by which the function f(x) is translated upward to give the function g(x). We can compare y-intercepts to see that the y-intercept of g(x) is 5 -2 = 3 units more than the y-intercept of f(x). This means ...

g(x) = f(x) +3

Comparing to ...

g(x) = f(x) +k

we see that k = 3.

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How to find the area r = 6 ft

Darya [45]

I am assuming this is for a circle??

Area for circle is pi*r^2

you didn't say what to use for pi so I will use 3.14

so area = 3.14 * 6^2

6^2 = 36

so 3.14*36 = 113.04 square feet. Round this answer off as needed.

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3 years ago

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Ivenika [448]

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3 years ago

The swimming pool at Spring Valley High School is a rectangle with a width of 65

tatiyna

Answer:

The area of the tiled floor is given by:

area = (190m)*w + 4*w^2

Step-by-step explanation:

Remember that the area of a rectangle of length L and width W is equal to:

A = W*L

For the case of the problem, we want to find the area of the tiled floor.

Then we need to find the difference between the area of the pool including the tiled floor and the area of the pool alone.

We know that the length of the tiled floor is w.

Then if the length of the pool is 30m, the length of the pool including the tiled floor is 30m + 2*w (because we have tiles in both sides)

And if the width of the pool is 65m, then the width of the pool including the tiled floor is 65m + 2*w

Now we want to find the difference between these areas, we have:

Area of the pool alone = 65m*30m = 1950m^2

Area of the pool including the tiles = (65m + 2*w)*(30m + 2*w)

The difference is:

D = (65m + 2*w)*(30m + 2*w) - 1950m^2

D = 65m*30m + 65m*2w + 30m*2w + 4*w^2 - 1950m^2

D = (65m + 30m)*2w + 4*w^2 + (65m*30m - 1950m^2)

D = 95m*2w + 4*w^2

D = (190m)*w + 4*w^2

This is the area of the tiled floor.

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

If the area of the circle below is 18 ft2, what is the area of the shaded sector?

iris [78.8K]

The answer is B. 4.5 ft2

8 0

3 years ago

Read 2 more answers