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The (greatest, least) of the common factors of 2 or more numbers is the (greatest, least) common factor of the numbers.

Liula [17]

Answer:

The greatest of the common factors of 2 or more numbers is the greatest common factor of the numbers.

Step-by-step explanation:

As we know that

The greatest common factor of two or more whole numbers is the largest whole number that divides evenly into each of the numbers.

Thus, the greatest of the common factors of 2 or more numbers is the greatest common factor of the numbers.

For example, lets find the greatest common factor of $36$ and $54$ .

The possible factors of $36$ are:

1, 2, 3, 4, 6, 9, 12, 18, and 36.

The possible factors of $54$ are:

1, 2, 3, 6, 9, 18, 27, and 54.

The common factors of 36 and 54 are:

$1, 2, 3, 6, 9, 18$

So, from the common factors, it is easy to figure out that 18 is the greatest common factor.

4 0

3 years ago

Find the least common denominator (LCD) of 10/7 and 1/6 .

Agata [3.3K]

Answer:

im not so positive sorry

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

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

A cup is filled with pencils. The teacher sharpen 1/4 of them. A student sharpen 2/3 of them. The fraction of pencils left to be

Bad White [126]

Answer:

1/12

Step-by-step explanation:

1/4+2/3=11/12

4 0

2 years ago

A family member has some five dollar bills and one dollar bills in their wallet. Altogether she has 18 bills and a total of $62.

Pepsi [2]

X - the number of five dollar bills
y - the number of one dollar bills

She has 18 bills in total.
$x+y=18 \\ y=18-x$

She has $62.
$5x+y=62 \\
y=62-5x$

Set both expressions equal to each other:
$18-x=62-5x \\
-x+5x=62-18 \\
4x=44 \\
x=11 \\ \\ y=18-x=18-11=7$

She has 11 five dollar bills and 7 one dollar bills.

7 0

3 years ago