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Lubov Fominskaja [6]
3 years ago
13

Keyoka has some bows. she has 8 more purple bows than green bows. she has twice as many orange bows as purple bows. she has 9 gr

een bows. how many bows does keyoka have altogether? student's solution: work backwards. keyoka has 9 green bows. she has 8 more purple bows than green bows (9 8 = 17). she has twice as many orange bows as purple bows (17 × 2 = 34). so keyoka has 9 17 34 or 60 bows. use a similar strategy to solve this problem. jessica has some markers. she has 5 more black markers than blue markers. she has 4 times as many red markers as black markers. she has 8 blue markers. how many markers does jessica have altogether?
Mathematics
2 answers:
valentinak56 [21]3 years ago
6 0
The answer is 73 markers.

Jessica has 8 blue markers. She has 5 more black markers than blue markers (5 + 8 = 13). She has <span>4 times as many red markers as black markers (13 * 4 = 52)</span>. So <span>Jessica </span>has 8 + 13 + 52 or 73 markers.
yarga [219]3 years ago
6 0

Jessica has 8 + 13 + 52 or 73 markers.

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8/5 ( 2x + 13 ) - 3 = 22
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Answer:

x =  \frac{21}{16} = 1 5/16

[/tex] 1 \frac{5}{16}[/tex]

Step-by-step explanation:

\frac{8}{5} (2x + 13) - 3 = 22

Expand

\frac{16}{5} x +  \frac{104}{5}  - 3 = 22 \\  \frac{16}{5} x +  \frac{89}{5}  = 2

Collect like terms and simplify

\frac{16}{5} x = 22 -  \frac{89}{5}  \\  \frac{16}{5} x =  \frac{21}{5}

Cross Multiply

80x = 105

Divide both sides of the equation by 80

\frac{80x}{80}  =  \frac{105}{80}  \\ x =  \frac{21}{16}

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3 years ago
An envelope is 60 centimeters wide. About how many inches wide is the envelope? (1 inch ≈ 2.5 centimeters)
Dima020 [189]

Answer:

60/2.5 = 24 cm

Step-by-step explanation:

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2 years ago
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Find the number of positive integers less than 100,000 whose digits are among 1, 2, 3, and 4. Hint consider the
Verdich [7]

Answer:

<em>1364</em> is the number of possibilities for positive integers less than 1,00,000.

Step-by-step explanation:

<em>1. 5 digit numbers:</em>

We have 5 places here, and each place can have 4 options because repetition is allowed.

So, total number of possibilities for 5 digit numbers:

4 \times 4 \times 4 \times 4 \times 4\\ \Rightarrow 1024

<em>2. 4 digit numbers:</em>

We have 4 places here, and each place can have 4 options because repetition is allowed.

So, total number of possibilities for 4 digit numbers:

4 \times 4 \times 4 \times 4 \\ \Rightarrow 256

<em>3. 3 digit numbers:</em>

We have 3 places here, and each place can have 4 options because repetition is allowed.

So, total number of possibilities for 3 digit numbers:

4 \times 4 \times 4 \\ \Rightarrow 64

<em>4. 2 digit numbers:</em>

We have <em>2</em> places here, and each place can have 4 options because repetition is allowed.

So, total number of possibilities for 2 digit numbers:

4 \times 4 \\ \Rightarrow 16

<em>5. 1 digit numbers:</em>

We have 1 place here, and each place can have 4 options because repetition is allowed.

So, total number of possibilities for 1 digit numbers:

4

We can add all the above possibilities to find the total.

So,  number of positive integers less than 100,000 whose digits are among 1, 2, 3, and 4 = 1024 + 256 + 64 + 16 + 4 = <em>1364</em>

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