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3 years ago

How many ounces are 3 pounds ?

Mathematics

2 answers:

Natalka [10]3 years ago

4 0

Answer: 48 ounces in 3 pounds (16 ounces to a pound)

guajiro [1.7K]3 years ago

3 0

Answer:

if you multiply the mass value by 16 you would get 48

3 times 16 equals 48

Step-by-step explanation:

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If the equation yx = zx is true, when yx is positive, x /= 0, and z is

denis23 [38]

Answer: negative

Step-by-step explanation:

With yx = zx, then by the Multiplicative Property of Equality, y = z. Since z is a negative integer, then y is also a negative integer. Since yx is positive, with y equal to a negative integer, then x is negative.

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2 years ago

Lucy's dog lost 6 pounds. How much weight does her dog need to gain in order to have a net change of 0 pounds?

nexus9112 [7]

Answer:

6 pounds

Step-by-step explanation:

if you lose 6 pounds and gain 6 pounds that make it a net change of 0 pounds

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3 years ago

Read 2 more answers

A landscape contractor planted 37 tulip bulbs, 53 daffodil bulbs, and 82 crocus bulbs in each field. He planted bulbs in 174 fie

ohaa [14]

He planted D. 29,928 Bulbs
First Add up all the bulbs.
Then multiply the # of bulbs with the # of fields.

3 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

The sides of a triangle have lengths 3, 25, and 27. What kind of triangle is it ?

Andrew [12]

Answer:

Step-by-step explanation:

it's obtuse the sides are to wide

3 0

2 years ago