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Rudiy27
2 years ago
12

Factor completely and then place the factors in the proper location on the grid.

Mathematics
1 answer:
Pie2 years ago
5 0
Answer is :
can find it on
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(X-2) is a factor of x^4+2x^3-7x^
zysi [14]

Answer:

x−2x4+2x3−7x2−8x+12=x3+4x2+x−6

The rational root theorem suggests that other possible roots may be -6, 6, -3, 3, -2, 2, -1, and 1. It turns out that x=-2x=−2 is a root, since (-2)^3+4(-2)^2+(-2)-6=0(−2)3+4(−2)2+(−2)−6=0 , so x+2x+2 is also a factor and we have

\dfrac{x^4+2x^3-7x^2-8x+12}{(x-2)(x+2)}=x^2+2x-3(x−2)(x+2)x4+2x3−7x2−8x+12=x2+2x−3

Finally, we can factorize the remaining quotient easily:

x^2+2x-3=(x+3)(x-1)x2+2x−3=(x+3)(x−1)

so the other factors are x+2x+2 , x+3x+3 , and x-1x−1 .

6 0
2 years ago
What is the equation of a hyperbola with a = 9 and c = 12? Assume that the transverse axis is horizontal
Lorico [155]
The general equation of a hyperbola with a horizontal transverse axis is defined as:
x²/a² - y²/b² = 1

Solving for b², we use the formula: a² + b² = c²
b² = 12² - 9² = 63

Equation of our hyperbola will be:
x²/81 - y²/63 = 1
4 0
3 years ago
LOTS OF POINTS!!<br><br> answer 4-6
riadik2000 [5.3K]

Answer:

Step-by-step explanation:

4 1x2x3 =6

5 3x4x1=12

6 2x3x2=12

Hope that help

4 0
3 years ago
Read 2 more answers
The oil painting is 90% off. it is now $17. what was the original price?
lutik1710 [3]

Answer:

Let us assume that the original purse is $100. The price after the first reduction is $80. After the second reduction the price is now $56.

Step-by-step explanation:

hope this helps

7 0
2 years ago
A radioactive substance has a continuous decay rate of 0.056 per minute. How many grams of a 120 gram sample will remain radioac
sammy [17]

Answer:

The mass of the radioactive sample after 40 minutes is 12.8 g.

Step-by-step explanation:

The mass of the sample can be found by using the exponential decay equation:

N_{t} = N_{0}e^{-\lambda t}

Where:

N(t): is the amount of the sample at time t =?

N₀: is the initial quantity of the sample = 120 g

t = 40 min

λ: is the decay constant =  0.056 min⁻¹

Hence, the mass of the sample after 40 min is:

N_{t} = N_{0}e^{-\lambda t} = 120g*e^{-0.056min^{-1}*40 min} = 12.8 g

     

Therefore, the mass of the radioactive sample after 40 minutes is 12.8 g.

I hope it helps you!

5 0
3 years ago
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