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kumpel [21]
2 years ago
13

Select the correct answer. Tracy is practicing singing for a competition. She increases the length of her daily practices by the

same amount each day and records her time in the table. Day Practice Time (minutes) 1 25 2 36 3 47 4 58 5 69 Select the explicit function which defines the sequence represented in the table.
Mathematics
1 answer:
Lostsunrise [7]2 years ago
7 0

Answer:

f(x) = 11x +14

Step-by-step explanation:

Given

The table in the question

Required

The explicit function

First, we calculate the slope of the table:

m = \frac{y_2 - y_1}{x_2 - x_1}

Where:

(x_1,y_1) = (1,25)

(x_1,y_1) = (3,47)

So:

m = \frac{47 - 25}{3 - 1}

m = \frac{22}{2}

m = 11

The equation is then calculated using:

y - y_1 = m(x - x_1)

So, we have:

y - 25 = 11(x - 1)

y - 25 = 11x - 11

y = 11x - 11+25

y = 11x +14

So, the function is:

f(x) = 11x +14

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Answer:

Casey's Function has the greater rate of change

Step-by-step explanation:

The graph shows a ROC of 5/1

Get a common denominator (4)

5/1 * 4/4 = 20/4

20/4 > 13/4

Casey's function has a greater ROC.

6 0
3 years ago
A basketball team scored 78 points in their championship win over their rival. The team scored 25 baskets worth 2 points and mad
solong [7]
25(2) + 7(1) + 3x = 78....x represents the number of 3 point shots
50 + 7 + 3x = 78
57 + 3x = 78
3x = 78 - 57
3x = 21
x = 21/3
x = 7....so there were 7 three-point shots made
3 0
3 years ago
Round 11,000 to the nearest 10,000
s2008m [1.1K]
10,000 is the answer
3 0
3 years ago
Read 2 more answers
The top and bottom margins of a poster are each 15 cm and the side margins are each 10 cm. If the area of printed material on th
12345 [234]

Answer:

the dimension of the poster = 90 cm length and 60 cm  width i.e 90 cm by 60 cm.

Step-by-step explanation:

From the given question.

Let p be the length of the of the printed material

Let q be the width of the of the printed material

Therefore pq = 2400 cm ²

q = \dfrac{2400 \ cm^2}{p}

To find the dimensions of the poster; we have:

the length of the poster to be p+30 and the width to be \dfrac{2400 \ cm^2}{p} + 20

The area of the printed material can now be:  A = (p+30)(\dfrac{2400 }{p} + 20)

=2400 +20 p +\dfrac{72000}{p}+600

Let differentiate with respect to p; we have

\dfrac{dA}{dp}= 20 - \dfrac{72000}{p^3}

Also;

\dfrac{d^2A}{dp^2}= \dfrac{144000}{p^3}

For the smallest area \dfrac{dA}{dp }=0

20 - \dfrac{72000}{p^2}=0

p^2 = \dfrac{72000}{20}

p² = 3600

p =√3600

p = 60

Since p = 60 ; replace p = 60 in the expression  q = \dfrac{2400 \ cm^2}{p}   to solve for q;

q = \dfrac{2400 \ cm^2}{p}

q = \dfrac{2400 \ cm^2}{60}

q = 40

Thus; the printed material has the length of 60 cm and the width of 40cm

the length of the poster = p+30 = 60 +30 = 90 cm

the width of the poster = \dfrac{2400 \ cm^2}{p} + 20 = \dfrac{2400 \ cm^2}{60} + 20  = 40 + 20 = 60

Hence; the dimension of the poster = 90 cm length and 60 cm  width i.e 90 cm by 60 cm.

4 0
3 years ago
Evaluate the expression. 1/2 x (4+8)
Elanso [62]

Answer:

Hey there!

1/2 x (4+8)

1/2 x (12)

6

Hope this helps :)

8 0
3 years ago
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