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nordsb [41]
2 years ago
5

What is the side length of the smaller square plate on which 36 CM chopstick can fit along a diagonal without any overhang

Mathematics
1 answer:
HACTEHA [7]2 years ago
6 0

Step-by-step explanation:

2

57

65

566

77788

66777

66788

67899

6789

78899

789999

88889

8888888

788889

888888999999999

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What is the first step needed to solve 3 over 4 multiplied by x minus 3 equals negative 18?
kodGreya [7K]
Minus 3 from the 3 and the 18
7 0
3 years ago
What are the odds of rolling 2 dice and getting the sum of 7?
Yuliya22 [10]

Answer:

1/6

Step-by-step explanation:

favorable over total outcomes

favorable is 6

total is 36

6/36 = 1/6

6 favorable outcomes are:

(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }

total of 36 outcomes are :

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

testbookcom

quora Praneethreddy Ravi Shetye

5 0
2 years ago
For the data shown in the scatter plot, which is the best estimate of r? -0.95, -0.55, 0.55, 0.95
Liula [17]
From the data shown in the scater plots, the points are scattered apart which means that r is not close to 1.

Also the data points increase to the right which means that r is positive.

Therefore, the estimate for r in the data shown in the scatter plot is 0.55
6 0
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
PLEASE HELP!! ILL MARK BRAINLYEST!!!
wlad13 [49]

Answer:

b. lol.

Step-by-step explanation:

yeah I'm almost positive

4 0
3 years ago
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