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ankoles [38]
3 years ago
15

Customary Units of Capacity Questions fill in the blanks.

Mathematics
2 answers:
ddd [48]3 years ago
7 0
2 qt = 8 c
4 gal = 16 qt
2 gal = 16 pt
48 c = 3 gal
7 gal = 56 pt
4 gal = 32 pt
3 qt = 12 c
2 gal = 8 qt
12 c = 3 qt
5 gal = 20 qt
10 qt = 20 pt
16 c = 4 qt
15 pt = 30 c
10 gal = 40 qt
18 pt = 9 qt
Alik [6]3 years ago
5 0
4qt=8c
4gal=16qt
8gal=16pt
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Please explain. 100 / (-5) =
svp [43]
That would be -20. a negative divided by a negative number is positive, positive divided by negative is negative bc even numbers of negatives make it positive. and ofc 20 x 5 = 100, hence the ‘20’
5 0
3 years ago
Read 2 more answers
The tens digit of a certain number is five more than the units digit. The sum of the digits is 9. Find the number.
Leokris [45]
<span>the tens digit = n
</span><span>the units digit = n-5

n + n-5 = 9
2n = 9+5
2n = 14
n = 7  </span>←  the tens digit

the units digit = n-5 = 7 - 5 = 2

<span>the number is 72</span>
4 0
3 years ago
Read 2 more answers
If we wanted to measure the capacity of the waterfall and how deep it was, what unit of measure would we use?
AleksAgata [21]
Probably dept and lenght
4 0
3 years ago
Two show your work questions
Svet_ta [14]

Answer:

x = 1; y = 1; z = 0

Step-by-step explanation:

First problem:

First equation minus second equation: 3y + 2z = 3      Eq. A

Second equation plus third equation: 2y + 4z = 2        Eq. B

-2 times Eq. A plus Eq. B: -4y = -4

y = 1

Substitute y = 1 in Eq. A: 3 + 2z = 3

2z = 0

z = 0

Substitute z = 0 and y = 1 in original first equation:

-2x + 2 + 0 = 0

-2x = -2

x = 1

Answer: x = 1; y = 1; z = 0



5 0
3 years ago
Use the value of the first integral I to evaluate the two given integrals. I = integral^3_0 (x^3 - 4x)dx = 9/4 A. integral^3_0 (
Troyanec [42]

Answer:

A) -9/2

B) 9/4

C) -9/2, same as A)

Step-by-step explanation:

We are given that I=\int_0^3 x^3-4x dx=9/4. We use the properties of integrals to write the new integrals in terms of I.

A) \int_0^3 8x-2x^3 dx=\int_0^3 -2(x^3-4x) dx=-2\int_0^3 x^3-4x dx=-2I=-9/2. We have used that ∫cf dx=c∫f dx.

B) \int_3^0 4x - x^3 dx=-\int_0^3 (4x-x^3) dx=\int_0^3-(4x-x^3) dx=\int_0^3 x^3-4x dx=I=9/4. Here we used that reversing the limits of integration changes the sign of the integral.

C) It's the same integral in A)

4 0
3 years ago
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