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2 years ago

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In the equation ax2 + bx + c = 0, if b is equal to the new equation is In solving this form of equation, we will use the princip

le of

1 answer:

Sphinxa [80]2 years ago

8 0

Answer:

The answer is b = ax- c/x

when i put the / it means fraction

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A rock thrown vertically upward from the surface of the moon at a velocity of 36m/sec reaches a height of s = 36t - 0.8 t^2 met

Verdich [7]

Answer:

a. The rock's velocity is $v(t)=36-1.6t \:{(m/s)}$ and the acceleration is $a(t)=-1.6 \:{(m/s^2)}$

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

Velocity is defined as the rate of change of position or the rate of displacement. $v(t)=\frac{ds}{dt}$
Acceleration is defined as the rate of change of velocity. $a(t)=\frac{dv}{dt}$

a.

The rock's velocity is the derivative of the height function $s(t) = 36t - 0.8 t^2$

$v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t$

The rock's acceleration is the derivative of the velocity function $v(t)=36-1.6t$

$a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6$

b. The rock will reach its highest point when the velocity becomes zero.

$v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5$

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

$s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405$

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = $\frac{405}{2} =202.5 \:m$

To find the time it reach half its maximum height, we need to solve

$36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41$

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

$36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45$

The rock is aloft for 45 seconds.

5 0

3 years ago

betty closes the nozzle and fills it completely with a liquid. She then opens the nozzle. If the liquid drips at the rate of 14

natta225 [31]

Answer:

4.71 minutes

Step-by-step explanation:

Incomplete question [See comment for complete question]

Given

Shape: Cone

$r = 3$ -- radius

$h = 7$ --- height

$Rate = 14in^3/min$

Required

Time to pass out all liquid

First, calculate the volume of the cone.

This is calculated as:

$V = \frac{1}{3} \pi r^2h$

This gives:

$V = \frac{1}{3} * 3.14 * 3^2 * 7$

$V = \frac{1}{3} * 197.82$

$V = 65.94in^3$

To calculate the time, we make use of the following rate formula.

$Rate = \frac{Volume}{Time}$

Make Time the subject

$Time= \frac{Volume}{Rate }$

This gives:

$Time= \frac{65.94in^3}{14in^3/min}$

$Time= \frac{65.94in^3}{14in^3}min$

Cancel out the units

$Time= \frac{65.94}{14} min$

$Time= 4.71 min\\$

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2 years ago

What expression is equivalent to 4,325,000,000

solmaris [256]

It’s B
Since 10^-6 means 6 0

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Alekssandra [29.7K]

33/40 is the experimental probability.

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3 years ago

Reasoning point F lies on EG and point M lies on EN. If F,E, and M are collinear, what must be true of these rays.

almond37 [142]

This means that all of the points are co-linear. This is because if EG is a segment that contains F, and EN is a segment that contains M, then there can be two different segments. However, for a point F on the first segment and a point M on a second segment be in the same line as an end point of one of the segments, the segments have to be co-linear. They overlap.

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