How many solutions exist for the given equation? 3x(x - 2) = 22 - x

Mathematics

1 answer:

Mila [183]3 years ago

5 0

The solution exist for the given equation are
x=-2
and
x=11/3

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A 5-meter ladder is leaning against the side of a house. The foot of the ladder is pulled away from the house at a rate of 0.4 m

Flura [38]

<h2>The top of the ladder is descending at 0.3 m/s.</h2>

Step-by-step explanation:

By Pythagoras theorem we know that

Hypotenuse² = Base² + Perpendicular²

h² = b² + p²

We have for ladder

h = 5 m

b = 3 m

5² = 3² + p²

p = 4 m

$\frac{db}{dt}=0.4m/s\\\\\frac{dh}{dt}=0$

Differentiating h² = b² + p² with respect to time

$2h\times \frac{dh}{dt}=2b\times \frac{db}{dt}+2p\times \frac{dp}{dt}\\\\5\times 0=3\times 0.4+4\times \frac{dp}{dt}\\\\\frac{dp}{dt}=-0.3m/s$

The top of the ladder is descending at 0.3 m/s.

3 0

3 years ago

How do measurements of time differ for events in a frame of reference that moves at 50% of the speed of light relative to us? At

ZanzabumX [31]

Answer with explanation:

Relation between Speed , Distance and time

Distance =Speed × Time

→It means Speed is inversely Proportional to time.

As distance will remain constant , in that frame of reference

If speed of light in a Medium

$=s=3 \times 10^8 \frac{\text{meter}}{\text{second}}$

Then, time taken to cross the medium = t seconds or hours or another unit of time.

Now, If speed of light is 50% of the speed of light relative to us

That is Speed of light in another medium

$=w=\frac{50}{100} \times 3 \times 10^8\\\\=1.5 \times 10^8 \frac{\text{meter}}{\text{second}}$

Then, time taken to cross the medium = 2t seconds or hours or another unit of time.

Using Unitary Method

$\rightarrow v_{1}\times t_{1}=v_{2} \times t_{2}\\\\1.\rightarrow 3 \times 10^8 \times t=\frac{50}{100} \times 3 \times 10^8 \times t_{1}\\\\t_{1}=2t\\\\2.\rightarrow 3 \times 10^8 \times t=\frac{99.5}{100} \times 3 \times 10^8 \times t_{2}\\\\t_{2}=\frac{1000t}{995}\\\\t_{2}=\frac{200t}{199}$