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2 years ago

Vertex of y = x³ + x² - 4x - 4

Mathematics

1 answer:

Firlakuza [10]2 years ago

5 0

Answer:

(-2,0)

Step-by-step explanation:

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Simplify. (-7y^2+4y+9)+(6y^2-5y+9)-(2y^2-9y-2)

Natali5045456 [20]

Combine like terms

-7y² + 6y² - 2y² = -3y²

4y + (-5y) - (-9y) = 4y - 5y + 9y = 8y

9 + 9 - (-2) = 9 + 9 + 2 = 20

-3y² + 8y + 20 is your answer

hope this helps

7 0

2 years ago

See Attached picture.

agasfer [191]

We use the Work formula to solve for the unknown in the problem which is W = F x d. First, we solve for the Net Force acting on the car. The Net Force is the summation of all forces acting on the object. For this case, we assume that Friction Force is negligible thus the Net Force is equal to:

F = mgsinα in terms of SI units and in terms of english units we have F = m(g/g₀)(sin α) where g₀ is the proportionality factor, 32.174 ft lb-m / lb-f s²

F = 2500 (32.174/32.174) (sin 12°) = 519.78 lb

W = Fd = 519.78 lb (400 ft) = 207912 ft - lb or 20800 ft-lb

5 0

2 years ago

Find the rate of commission to the nearest whole percent if the sale was $5,000 and

tatiyna

Answer:

the answer would be 16%, but if its rounded to nearest whole percent it would be 20

Step-by-step explanation:

6 0

2 years ago

Nine more than the quotient of b and 4

erma4kov [3.2K]

Answer:

b/4+9

Step-by-step explanation:

4 0

3 years ago

A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11

mixer [17]

Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're given that the tank initially holds $A(0)=100$ lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

$\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}$
$\implies A'(t)+\dfrac{11}{200+33t}A(t)=484$

Find the integrating factor:

$\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}$

Distribute $\mu(t)$ along both sides of the ODE:

$(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}$
$\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}$
$A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt$
$A(t)=22(200+33t)^{2/3}+C$

Since $A(0)=100$ , we get

$100=22(200)^{2/3}+C\implies C\approx-652.39$

so that the particular solution for $A(t)$ is

$A(t)=22(200+33t)^{2/3}-652.39$

The tank becomes full when the volume of solution in the tank at time $t$ is the same as the total volume of the tank:

$200+(44-11)t=500\implies 33t=300\implies t\approx9.09$

at which point the amount of salt in the solution would be

$A(9.09)\approx733.47\text{ lb}$

4 0

2 years ago

Vertex of y = x³ + x² - 4x - 4​

Vertex of y = x³ + x² - 4x - 4