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Black_prince [1.1K]
3 years ago
14

State the vertex as an ordered pair. y = -|3x +5|+7

Mathematics
1 answer:
Airida [17]3 years ago
8 0
In plain and short, what value of "x" makes the expression |3x + 5| to 0?

well, we can check by simply setting it to 0 and solving for x.

\bf 3x+5=0\implies 3x=-5\implies \boxed{x=-\cfrac{5}{3}}\\\\
-------------------------------\\\\
\textit{therefore, the vertex of }y = -|3x+5|\boxed{+7}\textit{ is at }\left(-\frac{5}{3}~,~7  \right)

notice, all you do is, get the x-value that makes the expression to 0, and then append the constant outside it as the y-value, and that's the vertex.
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What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400. *work needed*
pickupchik [31]
<h3>Answer: x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

Step-by-step explanation:

Given the polynomial equation x^4-41x^2=-400.

Adding 400 on both sides to get rid 400 from right side and set 0 on right side, we get

x^4-41x^2+400=-400+400.

x^4-41x^2+400=0.

Factoring by product sum rule.

We need product of 400 and sum upto -41.

We can see that 400 = -25 × -16 = 400 and -25-16 = -41.

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x^4-25x^2-16x^2+400=0

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(x^4-25x^2)+(-16x^2+400)=0

Factoring out GCF of each group, we get

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(x^2-25)(x^2-16) =0

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(x^2-25) = x^2-5^2= (x-5)(x+5) and

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

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(x-5)(x+5)(x-4)(x+4) =0

Applying zero product rule,

x-5=0

x+5=0

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x+4=0.

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<h3>x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>
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4000 + 12 343 x 2 =(A)
Vikentia [17]

Answer:

12,686

Step-by-step explanation:

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