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xz_007 [3.2K]
3 years ago
10

2) What are the possible zeros of the following equation? y = 3x4 + 4x² – 5x + 10

Mathematics
1 answer:
Nana76 [90]3 years ago
4 0

Answer:

Solution:

(4x2 - 2x) - (-5x2 - 8x)

= 4x2 - 2x + 5x2 + 8x.

= 4x2 + 5x2 - 2x + 8x.

= 9x2 + 6x.

= 3x(3x + 2).

Answer: 3x(3x + 2)

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What are the solution(s) of x2-4-0?
Elanso [62]

Answer:

For x^2 - 4 = 0, x  = 2, or x = - 2.

Step-by-step explanation:

Here, the given expression is :

x^2 - 4 = 0

Now, using the ALGEBRAIC IDENTITY:

a^2 - b^2 = (a-b)(a+b)

Comparing this with the above expression, we get

x^2 - 4 = 0  = x^2 - (2)^2 = 0\\\implies (x-2)(x+2) = 0

⇒Either (x-2) = 0 , or ( x + 2) = 0

So, if ( x- 2)   = 0 ⇒ x =  2

and if ( x + 2) = 0   ⇒ x = -2

Hence, for x^2 - 4 = 0, x  = 2, or x = - 2.

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3 years ago
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Answer: its b

Step-by-step explanation:

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kotykmax [81]

Answer:

362880

Step-by-step explanation:

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8 0
3 years ago
Help me pls i keep getting 3/2 but thats not the answer
Gemiola [76]

\huge\text{Hey there!}

\large\boxed{\mathsf{\dfrac{6x^2}{y^3}}}\\\\\large\boxed{= \mathsf{\dfrac{6(4^2)}{4^3}}}\\\\\large\boxed{= \mathsf{\dfrac{6(4\times4)}{4\times4\times4}}}\\\\\large\boxed{= \mathsf{\dfrac{96}{16\times4}}}}\\\\\large\boxed{= \mathsf{\dfrac{96}{64}}}\\\large\boxed{\approx \mathsf{\dfrac{96\div32}{64\div32}}}\\\large\boxed{\approx \mathsf{\dfrac{3}{2} \rightarrow \bf \dfrac{96}{64}}}

\huge\text{Thus, your answer is: \boxed{\mathsf{Option\ A. } \  \mathsf{\dfrac{96}{64}}}}\huge\checkmark

\huge\text{Good luck on your assignment \& enjoy}\\\huge\text{your day!}

~\frak{Amphitrite1040:)}

3 0
2 years ago
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How do you dilate a point on a graph?
rusak2 [61]

Answer:

If the point that you are dilating is directly above the point of dilation and you are dilating by 3, you take the distance from the point of dilation and the point you are dilating and you multiply it by 3. That is where you put your new point.

Step-by-step explanation:

i found this on google

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