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ollegr [7]
3 years ago
11

Factorise 10x²+19x+6 Show working out please.

Mathematics
1 answer:
stepladder [879]3 years ago
8 0

Answer:

(2x+3)(5x+2)

Step-by-step explanation:

10x²+19x+6

multiply to get 60

add to get 19

10x²+19x+6

(10x²+15x)(+4x+6)

factorise:

10x²+15x

5x(2x+3)

4x+6

2(2x+3)

(2x+3)(5x+2)

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C alternate exterior angles converse theorem

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2 years ago
|x-8|&lt;13<br> What is the solution
maks197457 [2]

Answer:

  • x = (-5, 21)

Step-by-step explanation:

  • |x - 8| < 13

1 .

  • x - 8 < 13
  • x < 8 + 13
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2.

  • x - 8 > -13
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<u>Combined answer</u>

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5 0
3 years ago
Does anyone know know to do this ?
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7 0
3 years ago
A hair salon completed a survey of 367 customers about satisfaction with service(Dissatisfied, Neutral, Satisfied, or Very Satis
mamaluj [8]

The results of the survey are shown in the table.

It's required to calculate the following probabilities:

a) A customer is neutral

The row labeled as 'Neutral' has a total of 79 customers out of 367 in total.

The required probability is:

p=\frac{79}{367}

b) A customer is a walk-in

The first column labeled as 'Walk-in' has a total of 106 customers out of 367. The required probability is

p=\frac{106}{367}

c) A customer is dissatisfied and a walk-in.

We have to cross the row Dissatisfied with the column Walk-in. The number of customers that have both categories is 22 out of 367.

The required probability is:

p=\frac{22}{367}

d) A customer is very satisfied given he saw a TV ad.

This is a conditional probability, where the total is 111 because is the given condition. The customers that have both characteristics are 33, thus the required probability is:

p=\frac{33}{111}=\frac{11}{37}

e) A customer is satisfied or referred.

Total satisfied = 139

Total referred = 150

With both features = 62

Total customers with one or both features: 139 + 150 - 62 = 227

We have to subtract the common feature because it was counted twice.

The required probability is:

p=\frac{227}{367}

6 0
1 year ago
6x+18y=18 find y and z
Masja [62]

6x + 18y = 18. Let us take x = 0

6(0) + 18y = 18

0 + 18y = 18

18y = 18

y = 18/18

y = 1

(0, 1) [x = 0, y = 1]

6x + 18y = 18. Let us take y = 0

6x + 18(0) = 18

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x = 18/6

x = 3

(3, 0) [x = 3, y =0]

Infinite solutions can be found for x and y

3 0
3 years ago
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