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Dimas [21]
2 years ago
14

Damian needs to buy 5 pints of juice but the store doesn't sell juice in pints, so damian buys 3 quarts of juice. does damian bu

y enough juice if not how much more does he have left or need to have enough?
Mathematics
2 answers:
vesna_86 [32]2 years ago
4 0
He had bought enough juice.
spin [16.1K]2 years ago
3 0

Answer:

Damain has bought enough juice.

Step-by-step explanation:

1 qt(quart) = 2 pints.

3 qt = 6 pints.

6 pints > 5 pints (6 pints <em>is bigger than </em>5 pints)

Therefore, Damain has bought enough juice.

Please mark as brainliest!

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Can you guys check my work pls?
alex41 [277]
Hey number 8. Be super careful, less than always look at that it should be 5n-3
3 0
3 years ago
Plz help me Find LM.
Ira Lisetskai [31]

Because LP and NP are the same measure, that means that MP is a bisector. It bisects side LN and it also bisects angle LMN. Where MP meets LN creates right angles. What we have then thus far is that angle LMP is congruent to angle NMP and that angle LPM is congruent to angle NPM and of course MP is congruent to itself by the reflexive property. Therefore, triangle LPM is congruent to triangle NMP and side LM is congruent to side NM by CPCTC. Side LM measures 11.

3 0
2 years ago
Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.
finlep [7]

Answer:

a) v = \frac{[L]}{[T]} = LT^{-1}

b) a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}

c) \int v dt = s(t) = [L]=L

d) \int a dt = v(t) = [L][T]^{-1}=LT^{-1}

e) \frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

v = \frac{ds}{dt}

a= \frac{dv}{dt}

Part a

If we do the dimensional analysis for v we got:

v = \frac{[L]}{[T]} = LT^{-1}

Part b

For the acceleration we can use the result obtained from part a and we got:

a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}

Part c

From definition if we do the integral of the velocity respect to t we got the position:

\int v dt = s(t)

And the dimensional analysis for the position is:

\int v dt = s(t) = [L]=L

Part d

The integral for the acceleration respect to the time is the velocity:

\int a dt = v(t)

And the dimensional analysis for the position is:

\int a dt = v(t) = [L][T]^{-1}=LT^{-1}

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

7 0
3 years ago
What is the value of the expression? 30-5×4+2
belka [17]
30-5×4+2
30-20+2
10+2=12
7 0
3 years ago
Read 2 more answers
The average score of students in Mr. Blake's science class was 73 with a standard deviation of 11, while the average score of st
Pani-rosa [81]

Answer:

Mr Blake's science class

Step-by-step explanation:

The standard deviation is a measure of variation.

It tells us how far each of the data set in the distribution are far from the mean of the distribution.

The average score of students in Mr. Blake's science class was 73 with a standard deviation of 11, while the average score of students in Mrs. Arnold's class was 75 with a standard deviation of 10.

Since 11>10, Mr Blake's science class is more variable than Mrs. Arnold's class.

3 0
2 years ago
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