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chubhunter [2.5K]
3 years ago
6

What is a zero of the function f(x) = 5x-20

Mathematics
1 answer:
Svetlanka [38]3 years ago
7 0
That is where the function equals 0
0=5x-20
add 20 both sides
20=5x
divide 5
4=x

the zero is at x=4
(4,0)
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There are 25 students' names in a hat. You choose 5 names. Three are
gizmo_the_mogwai [7]

Answer:

15

Step-by-step explanation:

You have to calculate the boys name using ratio expression.

3/5 × 25

= 15

5 0
3 years ago
Totally cellular charges $30 for a monthly plan, with an additional .20 per text message. Gnarly Cells charges $40 for a monthly
Taya2010 [7]

Answer:

100 texts

Step-by-step explanation:

Let the amount of text messages be x. Since we are looking at making both plans equal, we add the fixed monthly charges to the number of messages multiplied by the cost.

The total amount charged by Totally will be 30 + 0.2x

The total amount charged by Gnarly will be

40 + 0.1x

Now since we are trying to get the number of text messages that make both charges equal, we equate the two charges and that is:

30 + 0.2x = 40 + 0.1x

40 - 30 = 0.2x - 0.1x

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x = 10/0.1 = 100 texts

5 0
3 years ago
Simplify -7.39 -(-4.8)- 5.32
PtichkaEL [24]

simplify -7.39 -(-4.8)- 5.32 = -7.91

7 0
3 years ago
Read 2 more answers
If I place $2,000 in my retirement account at the beginning of each year for 10 years. The account will earn 5 percent interest
Norma-Jean [14]

Answer:

poooooooooooop

Step-by-step explanation:

6 0
2 years ago
A binary operation is defined on the set of real numbers ℝ by
zloy xaker [14]

Answer:

I. m = 2401

II. ((n+1) ∆ y)/n = 1/n[(n – y + 2)(n – y) + 1]

Step-by-step explanation:

I. Determination of m

x ∆ y = x² − 2xy + y²

2 ∆ − 5 = √m

2² − 2(2 × –5) + (–5)² = √m

4 – 2(–10) + 25 = √m

4 + 20 + 25 = √m

49 = √m

Take the square of both side

49² = m

2401 = m

m = 2401

II. Simplify ((n+1) ∆ y)/n

We'll begin by obtaining (n+1) ∆ y. This can be obtained as follow:

x ∆ y = x² − 2xy + y²

(n+1) ∆ y = (n+1)² – 2(n+1)y + y²

(n+1) ∆ y = n² + 2n + 1 – 2ny – 2y + y²

(n+1) ∆ y = n² + 2n – 2ny – 2y + y² + 1

(n+1) ∆ y = n² – 2ny + y² + 2n – 2y + 1

(n+1) ∆ y = n² – ny – ny + y² + 2n – 2y + 1

(n+1) ∆ y = n(n – y) – y(n – y) + 2(n – y) + 1

(n+1) ∆ y = (n – y + 2)(n – y) + 1

((n+1) ∆ y)/n = [(n – y + 2)(n – y) + 1] / n

((n+1) ∆ y)/n = 1/n[(n – y + 2)(n – y) + 1]

7 0
3 years ago
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