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tester [92]
3 years ago
11

Drag each value or expression to the correct location on the equations and sentences. Each value and expression can be used more

than once, but not all values and expressions will be used.
Gary works at a bakery, and he needs to select a rectangular cookie sheet with an area of 192 square inches. The area of the cookie sheet is represented by the expression given below.

Complete the given statements, and find the width of the cookie sheet.

x^2+4x
Step 1: x^2+4x=192
Step 2: x(x+__)=192 Length is X. The width is__.
Step 3: x^2+4x-192=0
Step 4: (x+__)(x-12)=0
Because the length can't be__, the length is 12 and the width is __.

Options: 12, 4, x, x+4, -16, 16, x+12, x^2

Mathematics
2 answers:
sveticcg [70]3 years ago
8 0

Answer: The width of the rectangle will be 16 unit.

Explanation: Since, Given- 192 square unit is the area of the rectangle.

but again, according to the the area is represent by the expression x^2+4x

Then, Steps will be- Step 1:x^2+4x= 192

Step2:x(x+4) = 192, where x is the length of the rectangle. And the width is x+4.

Step3:  x^2+4x-192=0

Step4: (x+16)(x-12)=0

Because the lenght can not be negative, the lenght is 12 and width is 16.

so, option 4th will be right.

Nadusha1986 [10]3 years ago
7 0

That is the answer to your question.

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47

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I need to know unknown value 9 is 25 % of what number
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25% = 1/4

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4 years ago
a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6?6x7x7=294 b) How many three-digit numbers
love history [14]

Answer:

a) 294

b) 180

c) 75

d) 168

e) 105

Step-by-step explanation:

Given the numbers 0, 1, 2, 3, 4, 5 and 6.

Part A)

How many 3 digit numbers can be formed ?

Solution:

Here we have 3 spaces for the digits.

Unit's place, ten's place and hundred's place.

For unit's place, any of the numbers can be used i.e. 7 options.

For ten's place, any of the numbers can be used i.e. 7 options.

For hundred's place, 0 can not be used (because if 0 is used here, the number will become 2 digit) i.e. 6 options.

Total number of ways = 7 \times 7 \times 6 = <em>294 </em>

<em></em>

<em>Part B:</em>

How many 3 digit numbers can be formed if repetition not allowed?

Solution:

Here we have 3 spaces for the digits.

Unit's place, ten's place and hundred's place.

For hundred's place, 0 can not be used (because if 0 is used here, the number will become 2 digit) i.e. 6 options.

Now, one digit used, So For unit's place, any of the numbers can be used i.e. 6 options.

Now, 2 digits used, so For ten's place, any of the numbers can be used i.e. 5 options.

Total number of ways = 6 \times 6 \times 5 = <em>180</em>

<em></em>

<em>Part C)</em>

How many odd numbers if each digit used only once ?

Solution:

For a number to be odd, the last digit must be odd i.e. unit's place can have only one of the digits from 1, 3 and 5.

Number of options for unit's place = 3

Now, one digit used and 0 can not be at hundred's place So For hundred's place, any of the numbers can be used i.e. 5 options.

Now, 2 digits used, so For ten's place, any of the numbers can be used i.e. 5 options.

Total number of ways = 3 \times 5 \times 5 = <em>75</em>

<em></em>

<em>Part d)</em>

How many numbers greater than 330 ?

Case 1: 4, 5 or 6 at hundred's place

Number of options for hundred's place = 3

Number of options for ten's place = 7

Number of options for unit's place = 7

Total number of ways = 3 \times 7 \times 7 = 147

Case 2: 3 at hundred's place

Number of options for hundred's place = 1

Number of options for ten's place = 3 (4, 5, 6)

Number of options for unit's place = 7

Total number of ways = 1 \times 3 \times 7 = 21

Total number of required ways = 147 + 21 = <em>168</em>

<em></em>

<em>Part e)</em>

Case 1: 4, 5 or 6 at hundred's place

Number of options for hundred's place = 3

Number of options for ten's place = 6

Number of options for unit's place = 5

Total number of ways = 3 \times 6 \times 5 = 90

Case 2: 3 at hundred's place

Number of options for hundred's place = 1

Number of options for ten's place = 3 (4, 5, 6)

Number of options for unit's place = 5

Total number of ways = 1 \times 3 \times 5 = 15

Total number of required ways = 90 + 15 = <em>105</em>

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7 0
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