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3 years ago

11

Compare and Contrast how you would factor x^2−9 and x^2+9. For full credit, make sure you explain why they are factored differen

tly.

1 answer:

marishachu [46]3 years ago

7 0

The factored form of the expression $x^2 - 9$ is $(x+3)(x-3)$

The expression x^2 + 9 has no real roots hence cannot be factored

Quadratic equations are equations with a degree of 2.

Given the quadratic function $x^2-9 \ and \ x^2+9$

For the function $x^2 - 9$

$=x^2-9\\=x^2-3^2$

Using the difference of two squares:

$a^2-b^2=(a-b)(a+b)\\x^2-3^2=(x-3)(x+3)$

For the expression x^2 + 9

The expression x^2 + 9 has no real roots hence cannot be factored

Learn more on factorization here: brainly.com/question/20293447

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-37 - -47<br> gtfrtgyhbtgfvbnhjmrybgtsfvdcsx

vesna_86 [32]

Answer:

-37- -47=-10

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers

a total of 1500 raffle tickets were collected during the school fair. if you have 30 raffle tickets, what is the probability tha

hammer [34]

Answer: 1/50 or 0.02

Step-by-step explanation:

30 / 1500 = 1/50

1/50 or 0.02

3 0

3 years ago

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of <em>P</em> lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for <em>S</em>.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for <em>C</em> :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick <em>P</em> such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

5 0

3 years ago

Humberly has $24 and they want to buy at least 8 containers of yogurt. Yogurt is packaged in both large and small containers. La

yanalaym [24]

Answer:

Hence he will be 4 large containers and 4 small containers

Step-by-step explanation:

Given data

Let the number of small containers be x

and the number of large containers be y

x+y= 8---------1

also

2x+4y= 24-----2

the system of equation to solve the problem is

x+y= 8

2x+4y= 24

from 1

x=8-y

put this in 2

2(8-y)+4y= 24

16-2y+4y= 24

2y= 24-16

2y= 8

y= 8/2

y= 4

put y= 4 in 1

x+4=8

x= 8-4

x= 4

Hence he will be 4 large containers and 4 small containers

6 0

3 years ago

Gwar [14]

Answer:

Integers are always closed under subtraction, addition and multiplication, not division though. Hope this helped! :D

Step-by-step explanation:

4 0

3 years ago