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3 years ago

Solve for y. 2y - x = 3 4 (-y + 1)

Mathematics

2 answers:

Mars2501 [29]3 years ago

6 0

2y-x=3/4(-y+1)first distribute

2y-x=-3/4y+3/4 get the y on one side

2y-x+3/4y=3/4 add 3/4y to both sides

2y+3/4y=3/4+x add x to both sides

11/4y=3/4+x now divide by 11/4 (both sides)

y=3/11+4/11x

y=3+4x/11

choice c is the answer

AlekseyPX3 years ago

5 0

Answer: $\bold{C.\quad y=\dfrac{4x+3}{11}}$

<u>Step-by-step explanation:</u>

$2y-x=\dfrac{3}{4}(-y+1)\\\\\\\text{Multiply both sides by 4 to clear the denominator:}\\4(2y-x)=3(-y+1)\\\\\text{Distribute:}\\8y-4x=-3y+3\\\\\text{Add 3y to both sides:}\\11y-4x=3\\\\\text{Add 4x to both sides:}\\11y=4x+3\\\\\text{Divide both sides by 11:}\\\\\boxed{y=\dfrac{4x+3}{11}}$

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anzhelika [568]

Answer:

Yes, it is effectively infinite

Step-by-step explanation:

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3 years ago

Find the value of each variable.

Alex

Answer:

a. a×3= -30

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3 years ago

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To complete a task in 16 days a company needs

o-na [289]

Answer:

15 days

Step-by-step explanation:

Total hours required =

16×5×6 = 480

Let n be the no. of days

480 = n×8×4

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4 years ago

The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 – 5t2, where t is the tim

Leno4ka [110]

<h2>Time Interval : 1 sec to 1.414 sec</h2>

Step-by-step explanation:

The height of the stuntperson jumping off a $20\text{ }m$ is modelled as $h\text{ }=\text{ }20\text{ }-\text{ }5t^{2}$ . Here $t$ is the time in seconds.

We have to record the person when he is between the heights $15\text{ }m\text{ and }10\text{ }m$ from the ground.

When $h\text{ }=\text{ }15$ ,

$15\text{ }=20\text{ }-\text{ }5t^{2}\\5t^{2}\text{ }=\text{ }5\\t^{2}\text{ }=\text{ }1\\t\text{ }=\text{ }1\text{ }sec$

When $h\text{ }=\text{ }10$ ,

$10\text{ }=20\text{ }-\text{ }5t^{2}\\5t^{2}\text{ }=\text{ }10\\t^{2}\text{ }=\text{ }2\\t\text{ }=\text{ }\sqrt{2}\text{ }=\text{ }1.414\text{ }sec$

The stuntperson is at height $15\text{ }m$ at $t\text{ }=\text{ }1\text{ }sec$ , and is at height $10\text{ }m$ at $t\text{ }=\text{ }1.414\text{ }sec$ .

∴ The camera should film the person for the time interval $\textbf{1}\text{ }sec\text{ to }\textbf{1.414}\text{ }sec$ .

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3 years ago

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Bingel [31]

Answer:

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3 years ago