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2 years ago

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Express the product shown as a fraction in simplest form -3/7 dot -2/3 please help

1 answer:

attashe74 [19]2 years ago

8 0

The answer is $\frac{2}{7}$ .

Multiple numerators by nums, and denoms by denoms.

$\\\frac{-3}{7}$ · $\frac{-2}{3}$ = $\frac{6}{21}$

$\frac{6}{21}$ simplified is $\frac{2}{7}$

Hope this helped and branluest would be appreciated

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What is the value of x?<br><br> Enter your answer in the box.<br><br> x =

Dmitry_Shevchenko [17]

Answer: x = 20

We know opposite angles are equivalent so we can set both of these expressions equal to each other to solve for x.

3x + 50 = 6x - 10
50 = 3x - 10 (subtract 3x from both sides)
60 = 3x (add 10 to both sides)
20 = x (divide both sides by 3)

8 0

2 years ago

Please help soon! Thanks! Solve for x

spin [16.1K]

Answer:

$x=27\degree$

Step-by-step explanation:

$\because AC || DE.... (given)$

$\therefore m\angle CAD= m\angle EDF\\ (corresponding \: \angle s)$

$\therefore m\angle BAD= m\angle EDF\\(\because B\in \overleftrightarrow{AC})$

$\therefore m\angle BAD= 2x\\(\because m\angle EDF=2x)$

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5 0

2 years ago

What is this symbol called plz answer urgent

Svetlanka [38]

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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 fraction must be subtracted from the sum of 1/4 and 1/6 to have an average of 1/12 of all the two fractions

DanielleElmas [232]

Answer:

so 1/3 must be subtracted from the sum of 1/4 and 1/6 to have an average of 1/12 of all the two fractions.

Step-by-step explanation:

let the fraction be x

(1/4 + 1/6)-x = 1/12

or, 10/24 - x = 1/12

or, 5/12-1/12 = x

so, x = 4/12 = 1/3

8 0

2 years ago