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3 years ago

What are the solutions to the equation 3(x – 4)(x + 5) = 0?

Mathematics

1 answer:

iris [78.8K]3 years ago

4 0

Answer:

x=4 and x= - 5

Step-by-step explanation:

In order to solve this we should re-arrange the equation:

$(x-4)(x+5)=\frac{0}{3}$

This is equal to:

$(x-4)(x+5)=0$

Then we can seperate this into:

$(x-4)=0\\(x+5)=0$

So solving for x in both cases we get

$x-4=0 \\x=4\\And\\x+5=0\\x=-5\\$

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A culture started with 6,000 bacteria. After 5 hours, it grew to 7,800 bacteria. Predict how many bacteria will be present after

AleksandrR [38]

Answer:

After 5 hours, the amount of bacteria increased: 7800-6000=1800

After 1 hours, the amount of bacteria increased: 1800/5=360

After 18 hours, the amount of bacteria increased: 360*18 = 6480

=> Total amount of bacteria after 13 hours: 6000+6480 =12480

4 0

3 years ago

The length (in centimeters) of a typical Pacific halibut t years old is approximately f(t) = 200(1 − 0.956e−0.18t). Suppose a Pa

horsena [70]

Answer:

7.2 years

Step-by-step explanation:

$f(t) =200(1 - 0.956e^{-0.18t})$

where f(t) is the length and t is the number of years old

given the Mike measures 148 cm, we need to find out the age

So we plug in 148 for f(t) and solve for t

$148 =200(1 - 0.956e^{-0.18t})$

divide both sides by 200

$\frac{148}{200} = 1 - 0.956e^{-0.18t}$

Now subtract 1 from both sides

$-0.26= - 0.956e^{-0.18t}$

Divide both sides by -0.956

$\frac{0.26}{0.956} =e^{-0.18t}$

Now take ln on both sides

$\frac{0.26}{0.956} =e^{-0.18t}$

$ln(\frac{0.26}{0.956} )=-0.18tln(e)$

$ln(\frac{0.26}{0.956} )=-0.18t$

divide both sides by -0.18

t=7.2

So 7.2 years

5 0

3 years ago

Sam's take-home pay is $743. His deductions are $25 for OASDI, $5 for Medicare, and $27 for income tax. What is his gross pay?

notsponge [240]

The correct answer is $800

4 0

4 years ago

Read 2 more answers

Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the

Likurg_2 [28]

Answer:

(a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

Step-by step explanation:

Given X represents the number on die.

The possible outcomes of X are 1, 2, 3, 4, 5, 6.

For a fair die, $P(X)=\frac{1}{6}$

(a) Moment generating function can be written as $M_{x}(t)$ .

$M_x(t)=\sum_{x=1}^{6} P(X=x)$

$M_{x}(t)=\frac{1}{6} e^{t}+\frac{1}{6} e^{2 t}+\frac{1}{6} e^{3 t}+\frac{1}{6} e^{4 t}+\frac{1}{6} e^{5 t}+\frac{1}{6} e^{6 t}$

$M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) Now, find $E(X) \text { and } E\((X^{2})$ using moment generating function

$M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)$

$M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)$

$\Rightarrow E(X)=\frac{21}{6}$

$M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)$

$M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)$

$\Rightarrow E\left(X^{2}\right)=\frac{91}{6}$

Hence, (a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$ .

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

6 0

4 years ago

The movie theater received a shipment of candy boxes in a carton that has a volume of about 64,000 cubic inches. Each candy box

vova2212 [387]

Volumen carton= 64000 inches³
Volume of a candy box= 6 inx 5 inx 1/2
Volume of a candy box=15 inches³

Then, you have that the total candy boxes in the carton is:

x=640000 inches³/15 inches³
x=4267 candy boxes

About how many candy boxes are in the carton? There are 4267 candy boxes

4 0

3 years ago

Read 2 more answers