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3 years ago

(w+14)/(4w+6)=(3)/(4) show work plz

Mathematics

1 answer:

sergey [27]3 years ago

7 0

Answer:

w = -1.44 and w = -14.06

Step-by-step explanation:

That "(3) / (4)" in (w+14)/(4w+6)=(3)/(4) looks fishy. Assuming that you really meant 3/4, then we have (w+14)/(4w+6)=3/4.

Mult. both sides by 4 to remove the fraction: 4(w+14)/(4w+6) = 3.

Multiply out (w+14)/(4w+6): 4w^2 + 6w + 56w + 84. Combine the w terms, obtaining 4w^2 + 62w + 84. Then the original equation becomes:

4w^2 + 62w + 84 = 3, or 4w^2 + 62w + 81 = 0.

This is a quadratic equation, with a = 4, b = 62 and c = 81. The two roots (values of

w) are as follows:

-62 plus or minus √(62^2-4(4)(81)

----------------------------------------------------------

-62 plus or minus √(2548) -62 plus or minus 50.48

w = ------------------------------------------- = ---------------------------------------

8 8

Simplifying, we get w = -1.44 and w = -14.06

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Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

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Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$

= $\int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt$

= By using the Scientific calculator notation;

distance = 106.109 m

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3 years ago

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Answer:

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Step-by-step explanation:

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Murljashka [212]

Answer:

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Step-by-step explanation:

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To solve, lets find the volumes of all of the options...

V=l*w*h

A.
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B.
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C.
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D.
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We can conclude that C & D aren't the answer, since they contain prisms that don't have a volume of 72cm³.

Now lets solve for the surface area of A and B...

A.
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B.
sA=2(wh+lw+lh)=2(2*4+9*2+9*4)=2(8+18+36)=2(62)=124cm²
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Answer=A

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Answer:

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