PLEASE HELP ME ASAP I AM STUCK !!!!!!!!!!!Witch of the following are solutions to the equation below 5x^2-2x+16=4x^2+6x

Mathematics

1 answer:

poizon [28]4 years ago

7 0

X=4

Explanation

First subtract 4x^2 + 6x from both sides of the equation

X^2 - 8x + 16 = 0

Then let’s factorice the equation using the middle-term break:

x^2 - 4x - 4x + 16 = 0
x (x - 4) - 4 (x - 4) = 0
(x - 4) (x - 4) = 0
(x - 4) ^2 = 0
x = 0

Therefore the solution to the equation is x = 4

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If x and y are two distinct angles satisfying the equation Acos(z)+Bsin(z)=C

boyakko [2]

Answer:

sin(x+y) = $\frac{2AB}{(A^{2}+B^{2}) }$

Step-by-step explanation:

As x and y are angles satisfying the given equations ,

$Acos(x) + Bsin(x)= C\\Acos(y) + Bsin(y)= C$

Subtracting second equation from first,

$A(cos(x)-cos(y)) + B(sin(x)-sin(y)) =0$

Applying trigonometric formulas,

$2A(sin(\frac{x+y}{2})sin(\frac{y-x}{2} )) + 2B(cos(\frac{x+y}{2} )sin(\frac{x-y}{2} )) =0\\\\$

Cancelling 2 and $sin(\frac{x-y}{2} )$ ,

$Asin(\frac{x+y}{2} ) = Bcos(\frac{x+y}{2} )\\\\tan(\frac{x+y}{2} ) = \frac{B}{A}$

We know $sin(\alpha) = \frac{2tan(\alpha /2)}{1+tan^{2}(\alpha /2) }$

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4 0

3 years ago

Please help!

vfiekz [6]

Answer:

THE ANSWER IS 44!

Step-by-step explanation:

6 0

3 years ago

Which expression is a sum of cubes?

Katen [24]

Since $\sqrt[3]{32}\not\in\mathbb Z$ , you can omit B and C as possible solutions.

Now,

$-64x^6y^{12}+125x^{16}y^3=(-4x^2y^4)^3+(5x^{16/3}y)^3$

but since 16 is not divisible by 3, A is not correct either.

So D is the answer. To verify:

$64x^6y^{12}+125x^9y^3=(4x^2y^4)^3+(5x^3y)^3$