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3 years ago

What is the square root of 2/9?

Mathematics

1 answer:

fomenos3 years ago

4 0

Answer:

Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.

Exact Form:

√2/3

Decimal Form:

0.47140452

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A fulcrum moving a resistance of 200 g has a distance to the fulcrum of 20 cm, the effort mass of 50 g has a distance to the ful

S_A_V [24]

Answer:

The ideal mechanical advantage (IMA) is $4$ .

Step-by-step explanation:

The ideal mechanical advantage is the ratio of length of longer lever $L_e$ to that of shorter lever $L_r$

IMA $\frac{L_e}{L_r}$

Please refer to the image attached.

We could see that the the resistance load moves $10\ cm$ cm towards the fulcrum so the distance of resistance load from fulcrum $= (20-10) =10\ cm$

Now the as the effort force moves $40\ cm$ towards the fulcrum overall distance from the fulcrum to the effort force (load) $=(80-40)=40\ cm$

Plugging the values of the distances in IMA formula we can have.

IMA $=\frac{(80-40)}{(20-10)} =\frac{40}{10} =4$ .

So the IMA of the fulcrum (simple machine) $= 4$

3 0

4 years ago

What is the value of x that makes PQ←→||RS←→?

Inessa05 [86]

Answer:

x = 55

Step-by-step explanation:

For PQ and RS to be parallel then

∠ACQ = ∠RDB ( Alternate exterior angles ), thus

3x - 65 = 2x - 10 ( subtract 2x from both sides )

x - 65 = - 10 ( add 65 to both sides )

x = 55

3 0

3 years ago

What does y= -3x+3 equal

Andre45 [30]

Answer:

Solving for x, the answer is x = 1 - y/3

Hope this helps :D

7 0

3 years ago

Read 2 more answers

Analysis of an accident scene indicates a car was traveling at a velocity of 69.5 mph (31.1 m/s) along the positive x-axis at th

STatiana [176]

We can find the acceleration via

${v_f}^2-{v_i}^2=2a\Delta x$

We have

$\left(5.20\dfrac{\rm m}{\rm s}\right)^2-\left(31.1\dfrac{\rm m}{\rm s}\right)^2=2a(115\,\mathrm m)$

$\implies\boxed{a=-4.09\dfrac{\rm m}{\mathrm s^2}}$

Then by definition of average acceleration,

$a_{\rm ave}=\dfrac{v_f-v_i}t$

so that

$-4.09\dfrac{\rm m}{\mathrm s^2}=\dfrac{5.20\frac{\rm m}{\rm s}-31.1\frac{\rm m}{\rm s}}t$

$\implies\boxed{t=6.33\,\mathrm s}$

We alternatively could have found the time without knowing the acceleration. Since acceleration is constant, the average velocity is

$v_{\rm ave}=\dfrac{x_f-x_i}t=\dfrac{v_f+v_i}2$

Then

$\dfrac{115\,\rm m}t=\dfrac{5.20\frac{\rm m}{\rm s}+31.1\frac{\rm m}{\rm s}}2$

$\implies\boxed{t=6.33\,\mathrm s}$

7 0

3 years ago

-4.8=1.2s+2s+4<br><br><br>Please help me <br>

nikdorinn [45]

Simplify both sides of the equation
-4.8=1.2s+2s+4
Combine like terms
-4.8=(1.2s+2s)+(4)
-4.8=3.2s+4
Flip the equation
3.2s+4=-4.8
Subtract both sides by 4
3.2s+4-4=-4.8-4
3.2s=-8.8
Divide both sides by 3.2
3.2s/3.2= -8.8/3.2
S=-2.75 will be the answer hope it helps

6 0

3 years ago