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3 years ago

10

What is the equation for x=-5

1 answer:

Tatiana [17]3 years ago

6 0

X-5=x I'm sure it can be the correct

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dsp73

70 x 8.5% = 70 x 0.085 = 5.95

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3 years ago

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Use the interactive graph to plot each set of points. Which sets represent proportional relationships? Check all that apply.

dem82 [27]

Answer:

The table C is a linear functionFor a function to be a linear function, then the x and the y values must change at a constant rates.From the question, as the x values of table C constantly change by 1, the y values constantly change by -2This means that the table C is a linear function

Step-by-step explanation:

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1 year ago

Find the coordinates of the image after a reflection across x=3

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The correct answer is D

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3 years ago

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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.

Therefore,

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

Making it into two groups, we get

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

Factoring out GCF of each group, we get

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

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

Factoring out $(x^2-25)$ and $(x^2-16)$ separately by difference of the squares identity $a^2-b^2=(a-b)(a+b)$ , we get

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

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

Therefore,

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

Applying zero product rule,

x-5=0

x+5=0

x-4=0 and

x+4=0.

Therefore,

<h3>x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

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3 years ago

In x^a=y, y^b=z, z^c=x, then prove that abc=1

Marysya12 [62]

Step-by-step explanation:

Let, x^a =y......(1) and

y^b =z.....(2) and

z ^c =x......(3).

Now, using (1) in (2) we get,

x ^ab =z......(4).

Now, using (4) in (3) we get,

x ^abc =x

or, x ^abc =x ^1

or, abc=1.

Hope it will help :)

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2 years ago

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