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Zigmanuir [339]
2 years ago
12

Let f(t) give the number of liters of fuel oil burned in t days, and w(r) the liters burned in r weeks. Find a formula for w by

scaling the input of f.
Mathematics
1 answer:
viktelen [127]2 years ago
3 0

Answer:

7 f(t)

Step-by-step explanation:

So, our f(t) is the number of liters burned in t days. If t is 1, f(t)=f(1) and so on for every t.

w(r) id the number of liters in r weeks. This is, in one week there are w(1) liters burned.

As in one week there are 7 days, we can replace the r, that is a week, by something that represents 7 days. As 1 day is represented by t, one week can be 7t (in other words r = 7t). So, we have that the liters burned in one week are:

w(r) = w[7f(t)]

So, we represented the liters in one week by it measure of days.

So, we can post that the number of liters burned in 7 days is the same as the number of liters burned 1 day multiplied by 7 times. So:

w (r) = w[7 f(t)] = 7 f(t)

Here we hace the w function represented in terms of t instead of r.

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A solid machine part is to be manufactured as shown in the figure. The part is made by cutting a small cone off the top of a lar
Alexxx [7]

Answer:

The answer is 390 × pi.

Step-by-step explanation:

Big cone:

V = 1/3 x pi x r^2 x h

= 1/3 x 3.14 x 9^2 x 15

= 1272.3 cubic inches

Small cone:

V = 1/3 x pi x r^2 x h

= 1/3 x 3.14 x 3^2 x 5

= 47.1 cubic inches  

1272.3 - 47.1 = 1225.2 cubic inches

1272/3.14 ≈ 390

The answer is 390 × pi.

4 0
2 years ago
Find the values of x1 and x2 where the following two constraints intersect.
Arte-miy333 [17]

Answer: x1 = 251/26, x2 = -111/26

Step-by-step explanation:

Hi!

As you can see in the figure, the point you are looking for is the intersection of two lines.

The intersection point is found solving this system of linear equations (the point must satisfy both equations):

9x_1 +7x_2=57\\4x_1 + 6x_2 = 13

You can solve it, for example, by the method of substitution:

\text{solve for x1 in the first equation:}\\x_1 = \frac{1}{9}(57 - 7x_2)

Then plug x1 into equation 2, and solve for x2:

\frac{4}{9}(57-7x_2) + 6x_2 = 13\\\text{doing the algebra you get:}\\x_2 = \frac{-111}{26}

Then you use the value of x2 to get x1:

x_1 = \frac{1}{9}(57 - 7x_2)= \frac{1}{9}(57 + 7*\frac{111}{26}) = 251/26\\

4 0
2 years ago
A plumber charges a $45 fee to make a house call and then $25 for each hour of labor. The plumber uses an equation in the form e
sasho [114]
M=slope
In this case the slope would be $25
y=mx+b
y=25x+45

Answer: m=25
5 0
2 years ago
Read 2 more answers
NEED HELP FAST WILL GIVE BRAINIEST
vladimir1956 [14]

Answer:

x^2 + 8x - 65 = 0.

Step-by-step explanation:

In order to solve the side of the enlarged area of 81 square inches, we use the equation in solving the area of a square. Area = side^2.

the increase in side will be x, so (4+x)^2 = 81, x^2 + 4x + 4x + 16 = 81

x^2 + 8x + 16 - 81 = 0

x^2 + 8x - 65 = 0.

  Hoped this helped mark Brainliest!

5 0
2 years ago
Read 2 more answers
In a study of the progeny of rabbits, Fibonacci (ca. 1170-ca. 1240) encountered the sequence now bearing his name. The sequence
expeople1 [14]

The question in part c is not clear, nevertheless, part a and part b would be solved.

Answer:

a. The first twelve terms are:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.

b. The first ten terms are:

1.000, 1.000, 1.500, 1.667, 1.600, 1.625, 1.615, 1.619, 1.618, 1.618.

Step-by-step explanation:

a. Given

an + 2 = an + an + 1

where a1 = 1 and a2 = 1.

a3 = a1 + a2

= 2

a4 = a2 + a3

= 3

a5 = a3 + a4

= 5

a6 = a5 + a4

= 8

a7 = a6 + a5

= 13

a8 = a7 + a6

= 21

a9 = a8 + a7

= 34

a10 = a9 + a8

= 55

a11 = a10 + a9

= 89

a12 = a11 + a10

= 144

The first twelve terms are:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.

(b)

Given

bn = an+1/an

b1 = a2/a1

= 1/1 = 1.000

b2 = a3/a2

= 2/1 = 1.000

b3 = a4/a3

= 3/2 = 1.500

b4 = a5/a4

= 5/3 = 1.667

b5 = a6/a5

= 8/5 = 1.600

b6 = a7/a6

= 13/8 = 1.625

b7 = a8/a7

= 21/13 = 1.615

b8 = a9/a8

= 34/21 = 1.619

b9 = a10/a9

= 55/34 = 1.618

b10 = a11/a10

= 89/55 = 1.618

The first ten terms are:

1.000, 1.000, 1.500, 1.667, 1.600, 1.625, 1.615, 1.619, 1.618, 1.618.

6 0
2 years ago
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