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kolezko [41]
3 years ago
10

Question in image. :)

Mathematics
1 answer:
jarptica [38.1K]3 years ago
6 0

Answer:

All real numbers greater than or equal to 0.

General Formulas and Concepts:

<u>Algebra I</u>

  • Domain is the set of x-values that can be inputted into function f(x)

Step-by-step explanation:

According to the graph, we see that our x-values span from 0 to infinity. Since 0 is a closed dot, it is inclusive in the domain:

[0, ∞) or x ≥ 0 or All Real Numbers greater than or equal to 0.

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Find the value of x for which the lines l and m are parallel.
Rina8888 [55]

Answer:

A) 13

Step-by-step explanation:

l // m

9x + 10 = 127

9x = 117

x = 13

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Find the value of x and full segment. <br> V is the midpoint of LU. <br> UV = 6 - 3x, LU = 14x + 2
Angelina_Jolie [31]

Step-by-step explanation:

L ----------V-------------U

LV = VU

UV = 6-3x

LU = 14x + 2 = 2UV

14X +2 = 12 - 6X

20X = 10

X = 0.5

UV = 6 - 3(0.5)

= 4.5

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Dima020 [189]

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24

Step-by-step explanation:

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What is -1 ⅓ × -⅖<br><br>Please explain your answer and show your work.​
lakkis [162]

Answer:

\frac{8}{15}

Step-by-step explanation:

  1. Change the mixed number into an improper fraction: -1 × 3 = -3, -3 + 1 = -2. Put -2 over 3: \frac{-2}{3}  
  2. Multiply -2/3 and -2/5: \frac{-2}{3} × -\frac{2}{5} = \frac{8}{15}  

Therefore, the answer is \frac{8}{15}.

6 0
3 years ago
Let be the set of permutations of whose first term is a prime. If we choose a permutation at random from , what is the probabili
Dennis_Churaev [7]

Answer:

Pr = \frac{1}{6}

Step-by-step explanation:

Given

S = \{1,2,3,4,5\}

n = 5

Required

Probability the third term is 3

First, we calculate the possible set.

The first must be prime (i.e. 2, 3 and 5) --- 3 numbers

2nd \to 4\ numbers

3rd \to 3\ numbers

4th \to 2\ numbers

5th \to 1\ number

So, the number of set is:

S = 3 * 4 * 3 * 2 * 1

S = 72

Next, the number of sets if the third term must be 2

1st \to 2 i.e. 1 or 5

2nd \to 3\ numbers ---- i.e. remove the already selected first term and the 3rd the compulsory third term

3rd \to 1\ number i.e. the digit 2

4th \to 2\ numbers

5th \to 1\ number

So

r = 2 * 3 * 1 * 2 * 1

r = 12

So, the probability is:

Pr = \frac{r}{S}

Pr = \frac{12}{72}

Pr = \frac{1}{6}

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