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LUCKY_DIMON [66]
4 years ago
11

The sum of two consecutive odd integers is 9 less than three times the first odd integer. Find the larger integer.

Mathematics
1 answer:
butalik [34]4 years ago
4 0

Answer:13

Step-by-step explanation:

2x+11=3x

11=x

x+2=13

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Step 3: Using multi-step equations to solve problems a) You are using the equation 4(p - 7) = 44 to determine how many pictures
mrs_skeptik [129]

Answer:

p = 18

Explanation:

\hookrightarrow \sf 4(p - 7) = 44

\sf change \ sides

\hookrightarrow \sf p-7\right}=\frac{44}{4}

\sf  simplify

\hookrightarrow \sf p-7\right}= 11

\sf change \ sides

\hookrightarrow \sf p\right}= 11+7

\sf final \ answer

\hookrightarrow \sf p\right}= 18

3 0
3 years ago
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2gallon container of window cleaner costs $15.04. What is the price per cup?
jeyben [28]

Answer:0.47 cents per cup

Step-by-step explanation:

Since they’re are 32 cups in 2 gallons, we can divide $15.04 by 32 and get the answer

5 0
4 years ago
4.3 Mr Mokwebo 's two sons, Refentse and Tokelo have just passed grade 7 and grade 9 respectively. He will give them some money
Bumek [7]

The amount of money each of Mr Mokwebo 's two sons; Tokelo and Refentse will get is R116 and R95 respectively.

<h3>How to write and solve equation:</h3>

  • Amount needed to buy chocolate = R253
  • Tokelo's money = x
  • Refentse's money = x - 21

Total amount needed = Tokelo's money + Refentse's money

253 = x + (x - 21)

253 = x + x - 21

253 = 2x - 21

253 - 21 = 2x

232 = 2x

divide both sides by 2

x = 232/2

x = 116

Therefore,

Tokelo's money = x = R116

Refentse's money = x - 21

= 116 - 21

= R95

Learn more about equation:

brainly.com/question/2972832

#SPJ1

6 0
2 years ago
Help me pleaseeee (10 points)
Lina20 [59]

Answer:

1.2 m off the ground

7 0
3 years ago
Help pls I'll give 25 points
RideAnS [48]

Answer:

\text{C. }60

Step-by-step explanation:

<u>Question from image</u>:

"The digits 1, 2, 3, 4, and 5 can be formed to arrange 120 different numbers. How many of these numbers will have the digits 1 and 2 in increasing order? For example, 14352 and 51234 are two such numbers."

Let's start by taking a look with the number 1. There are four possible places 1 could be, because there needs to be space for the 2 after it.

Checkmarks mark where the 1 can be, the <em>x</em> marks where it cannot be.

\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{X}

Let's start with the first position:

\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are four places the 2 can be. For each of these four places, we can arrange the remaining 3 digits in 3!=6 ways. Therefore, there are 4\cdot 6=\boxed{24} possible numbers when 1 is the first digit of the number.

Continue this process with the remaining possible positions for 1.

Second position:

\underline{X}\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are three places the 2 can be, since the 2 must be behind the 1. For each of these three places, the remaining 3 digits can be arranged in 3!=6 ways. Therefore, there are 3\cdot 6=\boxed{18} possible numbers when 1 is the second digit of the number.

The pattern continues. Next there will be 2 places to place the 2. For each of these, there are 3!=6 ways to rearrange the remaining 3 digits for a total of 2\cdot 6=\boxed{12} possible numbers when 1 is the third digit of the number.

Lastly, when 1 is the fourth digit of the number, there is only 1 place the 2 can be. For this one place, there are still 3!=6 ways to rearrange the remaining three numbers. Therefore, there are 1\cdot 6=\boxed{6} possible numbers when 1 is the fourth digit of the number.

Thus, there are 24+18+12+6=\boxed{60} numbers that will have the digits 1 and 2 in increasing order, from a set of 120 five-digit numbers created by the digits 1 through 5, where no digit may be repeated.

5 0
3 years ago
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