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Dmitrij [34]
2 years ago
5

Which expression is not equivalent to −4x3+x2−6x+8?

Mathematics
1 answer:
Gekata [30.6K]2 years ago
4 0

The expression which is not equivalent to the provided expression is expression number 1, x²(-4x+1)-2(3x-4).

<h3>What is the equivalent expression?</h3>

Equivalent expressions are the expression whose result is equal to the original expression, but the way of representation is different.

The expression given in the problem is

f(x)=-4x³+x²-6x+8

The option given as,

  • (1) x² (-4x+1)-2(3x-4)
  • (2) x(-4x²- x + 6) + 8
  • (3) -4x³ + (x - 2)(x - 4)
  • (4) -4(x³ - 2) + x(x - 6)

From the given expression, if we take out <em>x </em>from the first three terms, it looks like option 2.

f(x)=-4x³+x²-6x+8

f(x)=x(-4x²+x-6)+8

From the given expression, if the last three terms factored, it looks like option 3.

f(x)=-4x³+x²-6x+8

f(x)=-4x³+x^2-4x-2x+8

f(x)=-4x³+x(x-4)-2(x-4)

f(x)=-4x³+(x-4)(x-2)

Rearrange the given expression, and make it looks like option 4.

f(x)=-4x³+x²-6x+8

f(x)=-4x²+8+x²-6x

f(x)=-4(x³-2)+x(x-6)

Thus, the expression which is not equivalent to the provided expression is expression number 1, x²(-4x+1)-2(3x-4).

Learn more about the equivalent expression here;

brainly.com/question/2972832

#SPJ1

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Answer:

QII

Step-by-step explanation:

We can use the acronym: ASTC or All Students Take Calculus.

The first letter of each word tells us details within a certain quadrant.

All trig functions in QI will be positive.

Only sine (and cosecant) will be positive in QII.

Only tangent (and cotangent) will be positive in QIII.

And only cosine (or secant) will be positive in QIV.

We know that tan(θ)<0. In other words, it's negative.

So, it our angle θ <em>cannot</em> be in QI or QIII.

We also know that cos(θ)<0.

So, this removes QIV, since in QIV, cosine is positive.

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3 0
3 years ago
diferenta a doua numere este 47 .daca se imparte lprimul numar la al doilea se obtine catul 3 si rest 7 .care sunt cele doua num
lapo4ka [179]
A- b = 47
a :b = 3 r 7

a = 47 + b
(47 + b) : b = 3 r 7
(47 + b) + 7= b x 3
47 + 7 = 3b - b
54 = 2b
b = 54 : 2
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a = 47 + b
a = 47 + 27 
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6 0
3 years ago
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