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Murljashka [212]
3 years ago
11

In the paper drive, 4 boys brought old papers to school as follows: Fred, 30 pounds; Albert, 40 pounds; Henry, 10 pounds; and Pe

ter, 40 pounds. What was the average number of pounds they brought?
A: 25 pounds
B: 30 pounds
C: 35 pounds
D: 40 pounds
E: none of the above
Mathematics
2 answers:
seraphim [82]3 years ago
5 0
B: 30 pounds is the correct answer
ki77a [65]3 years ago
4 0

Answer:

The average number of pounds they bought was 30 pounds.

Step-by-step explanation:

we know that

To find out the average number of pounds, add the total number of pounds for each boy and then divide by the number of boys.

so

therefore

The average number of pounds they bought was 30 pounds.

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What is (34/8-16/9)-14/9
Allisa [31]
(34/8 - 16/9) = 89/36
(89/36 - 14/9) = 11/12

ANSWER: 11/12
7 0
3 years ago
Show the distributive property when solving 6 times 9 I need a answer
dsp73
Probably too late but 6*6+6*3
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3 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
Quadrilateral PQRS is transformed by translating it 6 units to the right and then rotating it 90° clockwise
Elina [12.6K]

Answer:

(y, -x -6)

Step-by-step explanation:

<em>The coordinates of PQRS is not given, however we'll assume the coordinates of PQRS to be (x,y)</em>

<em />

The first transformation: 6 units right

(x,y) -> (x + b, y)

Where b = 6

So, the new coordinates will be

(x + 6, y)

The second transformation: 90° clockwise

(x,y) -> (y,-x)

So, the new coordinates will be

(x + 6, y)-> (y, -x -6)

<em>Hence, the new coordinates of PQRS is (y, -x - 6)</em>

7 0
3 years ago
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