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2 years ago

I need help. Plz! ???

Mathematics

2 answers:

Anarel [89]2 years ago

8 0

Answer:

Step-by-step explanation:

If the graph of a relationship is a line or a ray through the origin, then it is proportional. If it is a line or ray that does not pass through the origin, then it is not proportional. Also, if it is not linear, then it is not proportional.

In order to be proportional the graph must be a straight line that always passes through (0,0).

Troyanec [42]2 years ago

3 0

Answer: No

Step-by-step explanation: No, because a proportional relationship is a line that goes up at a constant rate, and goes through the origin, an example of a proportional relationship is a straight line starting from the origin ( or what we call the starting point)

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(3-1) + (1 - 2) =<br> what is the answer to this question? i don’t understand it.

Ganezh [65]

Answer:

Step-by-step explanation:

$(3 - 1) + (1 - 2)$

Order of operations is () ^ * / + -

$2 + ( - 1)$

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3 years ago

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kvasek [131]

Answer: right side = as x → ∞, y → -∞ & left side = as x → -∞, y → -∞

<u>Step-by-step explanation:</u>

To find end behavior, we need to evaluate 2 things: Sign of the leading coefficient and Degree of the function.

Sign: Determines the end behavior of the right side.

Positive: right side goes to positive infinity
Negative: right side goes to negative infinity

Degree: Determines the end behavior of the left side

Odd: left side is opposite of right side
Even: left side is same as right side

f(x) = -4x⁶ + x² - 52

1. Sign is negative so right side goes to negative infinity (as x → ∞, y → -∞)

2. Degree is even so left side is the same as right side so it also goes to negative infinity (as x → -∞, y → -∞)

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3 years ago

Solve 3x+4+7x+8+7x3+7x3

kodGreya [7K]

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3 years ago

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If a polynomial function f(x) has roots -8, 1, and 6i, what must also be a root of f(x)?

Naddik [55]

Answer:

it must also have the root : - 6i

Step-by-step explanation:

If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.

This is because in order to render a polynomial with Real coefficients, the binomial factor (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:

$(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2$

where the imaginary unit has disappeared, making the expression real.

So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)

Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.

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3 years ago

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Goshia [24]

Answer:

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