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Sonja [21]
3 years ago
15

What is the product ? (4y-3)(2y^2+3y-5)

Mathematics
2 answers:
DIA [1.3K]3 years ago
5 0

Answer:

=8y^3+6y^2−29y+15

Step-by-step explanation:

(4y−3)(2y^2+3y−5)

=(4y+−3)(2y^2+3y+−5)

=(4y)(2y^2)+(4y)(3y)+(4y)(−5)+(−3)(2y^2)+(−3)(3y)+(−3)(−5)

=8y^3+12y^2−20^y−6y^2−9y+15

=8y^3+6y^2−29y+15

yuradex [85]3 years ago
4 0

Answer:

8y^3 + 6y^2 - 29y + 15

Step-by-step explanation:

8y^3 + 12y^2 - 20y - 6y^2 - 9y + 15

then

8y^3 + 6y^2 - 29y + 15 is the product

Best regards

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Solve for x, 0 ≤ x ≤2π (2cosx-1)(2sinx+√3 ) = 0
vladimir1956 [14]

Answer:

x = π/3, x = 5π/3, x = 4π/3

Step-by-step explanation:

Let's split the given equation (2cosx-1)(2sinx+√3 ) = 0 into two parts, and solve each separately. These parts would be 2cos(x) - 1 = 0, and 2sin(x) + √3  = 0.

2\cos \left(x\right)-1=0,\\2\cos \left(x\right)=1,\\\cos \left(x\right)=\frac{1}{2}

Remember that the general solutions for cos(x) = 1/2 are x = π/3 + 2πn and x = 5π/3 + 2πn. In this case we are given the interval 0 ≤ x ≤2π, and therefore x = π/3, and x = 5π/3.

Similarly:

\:2\sin \left(x\right)+\sqrt{3}=0,\\2\sin \left(x\right)=-\sqrt{3},\\\sin \left(x\right)=-\frac{\sqrt{3}}{2}

The general solutions for sin(x) = - √3/2 are x = 4π/3 + 2πn and x = 5π/3 + 2πn. Therefore x = 4π/3 and x = 5π/3 in this case.

So we have x = π/3, x = 5π/3, and x = 4π/3 as our solutions.

5 0
3 years ago
i cant find my answer. You should really give your people the answer they are looking fo instead of giving sujestions.
Vesnalui [34]

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Did you ask a question? I just checked your profile to see if you posted any but i dont see anything except this

Step-by-step explanation:

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k0ka [10]

Answer:

I got y = 10

or

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or the answer is y = -5x since the 4 stays the same but the -5x changes

I feel like you have to replace with x with the numbers

Step-by-step explanation:

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3 years ago
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