All categories

3 years ago

How do I solve x(3x-17)=0 using the square root property

Mathematics

1 answer:

scZoUnD [109]3 years ago

8 0

Answer:

x = 0 or $x = \frac{17}{3}$

Step-by-step explanation:

We have the quadratic equation of variable x and we have to solve the equation using the square root property.

Now, we have x(3x - 17) = 0

⇒ 3x² - 17x = 0

⇒ $3(x^{2} - \frac{17}{3}x ) = 0$

⇒ $x^{2} - 2 \times (\frac{17}{6}) \times x + (\frac{17}{6} )^{2} = (\frac{17}{6} )^{2}$

⇒ $(x - \frac{17}{6})^{2} = (\frac{17}{6} )^{2}$

Now,square rooting both sides we get,

$x - \frac{17}{6} = \pm\frac{17}{6}$

Therefore, either x = 0 or $x = \frac{17}{3}$ (Answer)

You might be interested in

An interview service provider charges 35 dollars sign up fee and 40 dollars per month for an internet plan whitch representation

kvasek [131]

Answer:

35+40x=y

Step-by-step explanation:

3 0

3 years ago

NEED ANSWER ASAP !!

NikAS [45]

Answer:

the plant was planted 2 inches tall and grew 1.1 inches per week

Step-by-step explanation:

5 0

2 years ago

812.16 is what percent of 752?

finlep [7]

108 is the answer to your question

3 0

2 years ago

Read 2 more answers

A solid rectangular box with height 5 m and square base with side lengths 4 m is built using a lightweight material whose densit

elena55 [62]

Answer:

627200LJ

Step-by-step explanation:

Height of box=5 m

Side of square base=4 m

Volume of rectangular box= $l\times b\times h$

Using the formula

Volume of rectangular box= $4\times 4\times 5=80m^3$

Density of material= $800kg/m^3$

We know that

$Mass=Density\times volume$

Using the formula

Mass of rectangular box= $80\times 800=64000 kg$

Gravity= $g=9.8m/s^2$

Weight of rectangular box,F= $mg=64000\times 9.8 N$

Let L be the vertical distance traveled by box

Total work done = $W=F\times displacement$

Therefore,total work done against gravity = $64000\times 9.8 L$ =627200 L J

Hence, the box requires work against gravity=627200LJ

5 0

3 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago