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IceJOKER [234]
3 years ago
10

Can someone plz help me. I need it asap. Plz someone help me!!!!!!!!!

Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
5 0
Lisa used the most amount of sugar. .6 pounds. 
Tom used 13% so 4 pounds * .13 = .52 pounds
Jane used 12/100th of the sugar, so 12%. 4 pounds * .12 = .48 pounds
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How is perimeter measured?
Trava [24]

Answer:

c- linear units

Step-by-step explanation:

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7 0
3 years ago
Round to the nearest hundredth<br> 1.359.377
pashok25 [27]

Answer:

1.360.000

Step-by-step explanation:

I hope this is correct, but you take the number behind the nearest hundredth. If it's five and up you go up a number and if it's 4 or lower you go down

7 0
3 years ago
James bought a cake that weighs 3 and 1 over 4 pounds. How many ounces does the cake weigh? Show your work. (5 points)
kolezko [41]

The weight of James cake in ounces is 52 ounces and the Pint of water dispensed in 20 seconds is 24 pints

<h3>Weight</h3>

  • Weight of cake = 3 1/4 pounds

1 pound = 16 ounces

Number of ounces the cake weigh = 3 1/4 pounds × 16 ounces

= 13/4 × 16

= (208) / 4

= 52 ounces

  • Water dispensed per second = 0.15 gallons
  • Number of seconds = 20 seconds

1 gallon = 4 quarts

0.15 gallon = 0.6 quarts

1 quart = 2 pints

0.6 quart = 1.2 pints

Pint of water dispensed in 20 seconds = 1.2 × 20

= 24 pints of water

Learn more about weight:

brainly.com/question/229459

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8 0
2 years ago
Solve 2x-5=7<br><br><br>Pls tell me
velikii [3]

Answer: if solving for x....

x= 6

Step-by-step explanation:

2x-5=7

2x-5+5=7+5

2x-5+5=7+5

2x/2= 12/2

x= 6

7 0
3 years ago
Read 2 more answers
Ask your teacher find the limit. use l'hospital's rule where appropriate. if there is a more elementary method, consider using i
liberstina [14]

Answer:

  ln(5/3)

Step-by-step explanation:

The desired limit represents the logarithm of an indeterminate form, so L'Hopital's rule could be applied. However, the logarithm can be simplified to a form that is not indeterminate.

<h3>Limit</h3>

We can cancel factors of (x-1), which are what make the expression indeterminate at x=1. Then the limit can be evaluated directly by substituting x=1.

  \diplaystyle \lim\limits_{x\to1}{(\ln(x^5-1)-\ln(x^3-1))}=\lim\limits_{x\to1}\ln{\left(\dfrac{x^5-1}{x^3-1}\right)}\\\\=\lim\limits_{x\to1}\ln\left(\dfrac{x^4+x^3+x^2+x+1}{x^2+x+1}\right)=\ln{\dfrac{5}{3}}

8 0
1 year ago
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