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2 years ago

Use the quadratic formula to find the zeros. 3x^2=8x+1

Mathematics

1 answer:

iVinArrow [24]2 years ago

6 0

Answer:

x1 = 2.7863
x2 = - 0.11963

Step-by-step explanation:

Transform the quadratic to

3x^2 - 8x - 1 = 0

I don't see any obvious factors to this equation so we need to use the quadratic equation.

$\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a}$

a = 3
b = -8
c = - 1

$\text{x = }\dfrac{ 8 \pm \sqrt{(-8)^2 - 4*3*(-1) } }{2*3}$

$\text{x = }\dfrac{( 8) \pm \sqrt{64 +12 } }{6}$

$\text{x = }\dfrac{( 8) \pm \sqrt{76 } }{6}$

x = (8 +/- 8.7178 ) / 6

x1 = (16.87178)/6 = 2.7863

x2= (8 -8.7178)/6 = -0.1196

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Britney is trying out for the track team along with 24 other people nine people were running distance seven people were there to

ivolga24 [154]

Answer:

The 6-minute half-mile time is continuous data; the number of people trying out for the team is discrete data.

Step-by-step explanation:

3 0

1 year ago

A = 12 and b = 24 ,what Is the area of the pencil

dezoksy [38]

Their is not a lot of information but I think it is 12*24=288

6 0

2 years ago

Two welders worked a total of 46 h on a project. One welder made $34/h, while the other made $39/h. If the gross earnings of the

Ratling [72]

Answer:

25 and 21 hours respectively

Step-by-step explanation:

Let the number of hours worked by each welder be x and y respectively.

They worked a total of 46 hours. This means :

x + y = 46 hours.......(I)

Now, given their hourly charges, since we have the total amount of money realized, we can make an equation out of it. This means:

34x + 39y = 1669........(ii)

We then solve both simultaneously. From I, x = 46 -y

We can substitute this into ii

34(46 -y) + 39y = 1669

1564 -34y + 39y = 1669

5y = 1669 - 1564

5y = 105

y = 105/5 = 21

x = 46 - y

x = 46 - 21 = 25 hours

The numbers of hours worked by the welders are 25 and 21 respectively

4 0

2 years ago

Brainliest to right person

atroni [7]

Answer:

4 ÷ 6/9 is also equal to 4 x 9/6. This is because when you divide by a fraction, you change the division sign to multiply and reverse the numerator and the denominator of the fraction. For example, if x/y was a fraction, it would become y/x. THIS IS ONLY DURING DIVISION!

So, we now that 4 ÷ 6/9 is equal to 4 x 9/6. 4 x 9/6 = (4x9)/6 = 36/6 = 6.

7 0

3 years ago

NO LINKS OR FILES!

Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

2 years ago