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goldfiish [28.3K]
3 years ago
13

PLEASE HELP ASAP!!!!I'll mark brainliest to the correct answers

Mathematics
1 answer:
Korvikt [17]3 years ago
7 0

Answer:

Option 2

Step-by-step explanation:

2 + ¾(t) > 10

¾(t) > 8

t > 32/3

t > 10 ⅔

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after making a deposit Puja had $264 in her savings account she knows that if she added $26 to the mouth originally in the accou
alex41 [277]

Answer:

$106.00

Step-by-step explanation:

origional amount = x

x + 26 + x + 26 = 264

2x + 52 = 264

2x = 264 - 52

2x = 212

x = 212/2

  = 106




Hope this helps you! Have an amazing day :)



3 0
3 years ago
For a grade pls help
Jet001 [13]

Answer:

x = 7

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Work out the percentage change to 2 decimal places when a price of £70 is increased to £99.
TiliK225 [7]

Answer: that is the solution to the question

Step-by-step explanation:

6 0
3 years ago
At the bank, Derek made 7 withdrawals, each in the same amount. His brother, John, made 5 withdrawals, each in the same amount.
lozanna [386]

Answer:

x=12.50

y = 17.50

Step-by-step explanation:

x = Amount of one Derek withdrawal

then amount of John withdrawal =x+5

Total withdrawals of Derek = no of times x one time withdrawal

= 7(x) = 7x ... i

No of John withdrawal = no of times x one time withdrawal

= 5(x+5)  ... ii

Given that i and ii are equal

i.e. 7x =5(x+5)

7x = 5x+25

2x =25

x = 12.50 dollars

Y = 17.50 dollars

Checking part:

total derek withdrawal = 7(12.5) =87.50

Joh's withdrawal = 5(17.50) = 87.50

Since both are equal, our answers are right.

Solution:  Derek withdrew each time 12.50 dollars each for 7 times and John withdrew 17.50 dollars each for 5 times.

8 0
3 years ago
How much heat is required to raise the temperature of
Elina [12.6K]

Answer:

The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .

Step-by-step explanation:

Given as :

The mass of liquid water = 50 g

The initial temperature = T_1 = 15°c

The final temperature = T_2  = 100°c

The latent heat of vaporization of water = 2260.0 J/g

Let The amount of heat required to raise temperature = Q Joule

Now, From method

Heat = mass × latent heat × change in temperature

Or, Q = m × s × ΔT

or, Q =  m × s × ( T_2 - T_1 )

So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )

Or, Q =  50 g × 2260.0 J/g × 85°c

∴   Q = 9,605,000  joule

Or, Q =  9,605 × 10³ joule

Or, Q = 9605 kilo joule

Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer

4 0
3 years ago
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