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2 years ago

15

Can someone help me this is really confusing????? What is the solution set for the open sentence with the given replacement set?

1 answer:

IgorLugansk [536]2 years ago

6 0

The answer is:

3

w^2 - 1 = w + 5

3^2 - 1 = w + 5

3^2 = 3(3) = 9

9-1 = 8

8 = 3 + 5

8 = 8

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Zatra is scuba diving 77 meters below the surface of the water After explorng a few minutes, she ascends (goes up) 45 meters to

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New elevation = 122 meter below

Step-by-step explanation:

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2 years ago

Giving brainly to who ever answers first.

s2008m [1.1K]

Answer:

1/9 for left

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6 0

2 years ago

Read 2 more answers

A.) pinky bought 1 1/2 kg of apples and 5 1/4 kg of mangoes and 1 1/2 Kg of oranges. Find the total weight of fruits B.) If her

Natasha2012 [34]

Answer:

a) The total weight of fruits is $8\,\frac{1}{4}$ kilograms, b) The weight of the fruits left is $4\,\frac{1}{2}$ kilograms.

Step-by-step explanation:

a) The total weight of fruits ( $m_{T}$ ) is calculated by the following formula:

$m_{T} = m_{a} + m_{m}+m_{o}$

Where:

$m_{a}$ - Total weight of apples, measured in kilograms.

$m_{m}$ - Total weight of mangoes, measured in kilograms.

$m_{o}$ - Total weight of oranges, measured in kilograms.

If $m_{a} = 1\,\frac{1}{2} \,kg$ , $m_{m} = 5\,\frac{1}{4}\,kg$ and $m_{o} = 1\,\frac{1}{2}\,kg$ , then:

$m_{T} = 1\,\frac{1}{2}\,kg + 5\,\frac{1}{4}\,kg + 1\,\frac{1}{2}\,kg$

$m_{T} = \frac{6}{4}\,kg + \frac{21}{4}\,kg + \frac{6}{4}\,kg$

$m_{T} = \frac{33}{4}\,kg$

$m_{T} = 8\,\frac{1}{4}\,kg$

The total weight of fruits is $8\,\frac{1}{4}$ kilograms.

b) The weight eaten by her family is determined by the following expression:

$m_{E} = m_{a,e} + m_{m,e} + m_{o,e}$

Where:

$m_{a,e}$ - Eaten weight of apples, measured in kilograms.

$m_{m,e}$ - Eaten weight of mangoes, measured in kilograms.

$m_{o,e}$ - Eaten weight of oranges, measured in kilograms.

Given that $m_{a,e} = \frac{3}{4}\,kg$ , $m_{m,e} = 2\,\frac{1}{2}\,kg$ and $m_{o,e} = \frac{1}{2}\,kg$ , the weight eaten by her family is:

$m_{E} = \frac{3}{4}\,kg + 2\,\frac{1}{2}\,kg + \frac{1}{2}\,kg$

$m_{E} = \frac{3}{4}\,kg + \frac{10}{4}\,kg + \frac{2}{4}\,kg$

$m_{E} = \frac{15}{4}\,kg$

$m_{E} = 3\,\frac{3}{4}\,kg$

The weight of the fruits left is found by subtraction:

$m_{R} = m_{T}-m_{E}$

$m_{R} = 8\,\frac{1}{4} \,kg -3\,\frac{3}{4}\,kg$

$m_{R} = \frac{33}{4}\,kg-\frac{15}{4}\,kg$

$m_{R} = \frac{18}{4}\,kg$

$m_{R} = 4 \,\frac{1}{2}\,kg$

The weight of the fruits left is $4\,\frac{1}{2}$ kilograms.

3 0

2 years ago