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3 years ago

Divide the given polynomial by the given monomial 3 x square - 2 x divided by X

Mathematics

1 answer:

inna [77]3 years ago

7 0

Dividing (3x^2-2x) by x gives out 3x-2

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Can someone please help my math assignment is due in 4 hours and I have no idea how to do these 2 problems

Nitella [24]

Answer:

hmmm

Step-by-step explanation:

I think its 3-5

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3 years ago

I NEED HELP PLS

S_A_V [24]

The answer is e I think

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3 years ago

Monica has $20. She needs to buy a gallon of milk that costs $2.50. She also wants to buy yogurt, which costs $1.20

Ket [755]

Answer:

$2.50 + $1.20x $\leq$ $20

Step-by-step explanation:

Monica needs to buy a gallon of milk ($2.50), but she also wants to buy yogurt which is $1.20 each. However, her total cost has to be under $20, and she doesn't know how much yogurt she can get. So the number of cups of yogurt is unknown, so it is represented by the variable x. Now, the milk and the unknown number of cups of yogurt must be less than or equal to $20.

5 0

3 years ago

HELP ASAP 20 POINTS CAUSE IM DESPERATE ToT

Elenna [48]

1. By de Moivre's theorem,

$\left(4\left(\cos\left(\dfrac{7\pi}9\right) + i \sin\left(\dfrac{7\pi}9\right)\right)\right)^3 = 4^3 \left(\cos\left(\dfrac{21\pi}9\right) + i \sin\left(\dfrac{21\pi}9\right)\right) \\\\ ~~~~~~~~ = 64 \left(\cos\left(\dfrac{7\pi}3\right) + i \sin\left(\dfrac{7\pi}3\right)\right) \\\\ ~~~~~~~~ = 64 \left(\cos\left(\dfrac\pi3\right) + i \sin\left(\dfrac\pi3\right)\right) \\\\ ~~~~~~~~ = 64 \left(\dfrac12 + i\dfrac{\sqrt3}2\right) \\\\ ~~~~~~~~ = \boxed{32 + 32\sqrt3\,i}$

2. First write the given number in exponential/trigonometric form.

$z = 5-5\sqrt3\,i$

has modulus

$|z| = \sqrt{5^2 + \left(-5\sqrt3\right)^2} = \sqrt{100} = 10$

and since it lies in the second quadrant of the complex plane, its argument is

$\arg(z) = \pi + \tan^{-1}\left(-\dfrac{5\sqrt3}5\right) = \pi + \tan^{-1}\left(-\sqrt3\right) = \pi - \dfrac\pi3 = \dfrac{2\pi}3$

So, we have

$z = 5 - 5\sqrt3\,i = 10 e^{i2\pi/3} = 10 \left(\cos\left(\dfrac{2\pi}3\right) + i \sin\left(\dfrac{2\pi}3\right)\right)$

Now we apply de Moivre's theorem again, and make sure to account for the multivalued-ness of the exponential function. For $k\in\{0,1,2,3,4\}$ , the fifth roots of $z$ are

$z^{1/5} = 10^{1/5} e^{i(2\pi/3 + 2\pi k)/5}$

$k=0 \implies z^{1/5} = 10^{1/5} e^{i2\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{2\pi}{15}\right) + i \sin\left(\dfrac{2\pi}{15}\right)\right)}$

$k=1 \implies z^{1/5} = 10^{1/5} e^{i8\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{8\pi}{15}\right) + i \sin\left(\dfrac{8\pi}{15}\right)\right)}$

$k=2 \implies z^{1/5} = 10^{1/5} e^{i14\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{14\pi}{15}\right) + i \sin\left(\dfrac{14\pi}{15}\right)\right)}$

$k=3 \implies z^{1/5} = 10^{1/5} e^{i20\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{4\pi}3\right) + i \sin\left(\dfrac{4\pi}3\right)\right)}$ $k=4 \implies z^{1/5} = 10^{1/5} e^{i26\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{26\pi}{15}\right) + i \sin\left(\dfrac{26\pi}{15}\right)\right)}$

6 0

2 years ago

What is (3x9-6+3\10)

aliya0001 [1]

The order of operations says:
P arentheses
E xponents
M ultiplication
D ivision
A ddition
S ubtraction

So first we will do 3 x 9 = 27. (27 - 6 + 3 / 10)
Next do 3/10 = .3. (27 - 6 + .3)
Then solve 27-6 = 21. (21 + .3)
Lastly do the last part! The answer is 21.3, hope this helps!

4 0

3 years ago

Divide the given polynomial by the given monomial 3 x square - 2 x divided by X ​

Divide the given polynomial by the given monomial 3 x square - 2 x divided by X