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LUCKY_DIMON [66]

2 years ago

11

Can two polynomials be subtracted such that the difference is a trinomial? If so, give an example. (1 point) O Yes. (3x + 3) - (

3x + 2y + 3) O Yes. (3x + 3) - (2x + y + 3) O No. Two polynomials can never be subtracted such that the difference is a trinomial. O Yes. (3x + 3) - (2x + 2y + 4)

1 answer:

Paha777 [63]2 years ago

4 0

Answer:

(d) Yes. (3x + 3) - (2x + 2y + 4)

Step-by-step explanation:

The first answer choice results in a monomial, -2y

The second answer choice results in a binomial, x -y

The third answer choice is proved wrong by the fourth answer choice.

The fourth answer choice results in x -2y -1, a trinomial.

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2 years ago

Give the null and alternative hypotheses in symbolic form that would be used in a hypothesis test of the following claim:

irakobra [83]

Answer:

Step-by-step explanation:

The null hypothesis is usually the default statement. The alternative is the opposite of the null and usually tested against the null hypothesis

In this case study,

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3 years ago

Jerome found the lengths of each side of triangle QRS as shown, but did not simplify his answers. Simplify the lengths of each s

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Answer:

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Step-by-step explanation:

option D for Edgu hope this helps :)

9 0

2 years ago

Read 2 more answers

Define the type of sequence below.<br> -2, 0, 2, 4, 6,

mihalych1998 [28]

Answer:

Arithmetic sequence.

Step-by-step explanation:

We have been given a sequence : -2, 0, 2, 4, 6. We are asked to define the type of our given sequence.

We can see from our given sequence that each next term is 2 more than the previous term of the sequence.

$0--2=0+2=2$

$2-0=2$

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3 0

2 years ago

Read 2 more answers

Suppose a, b denotes of the quadratic polynomial x² + 20x - 2022 & c, d are roots of x² - 20x + 2022 then the value of ac(a

Alja [10]

<h3><u>Correct Question :- </u></h3>

$\sf\:a,b \: are \: the \: roots \: of \: {x}^{2} + 20x - 2020 = 0 \: and \: \\ \sf \: c,d \: are \: the \: roots \: of \: {x}^{2} - 20x + 2020 = 0 \: then \:$

$\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) =$

(a) 0

(b) 8000

(c) 8080

(d) 16000

$\large\underline{\sf{Solution-}}$

Given that

$\red{\rm :\longmapsto\:a,b \: are \: the \: roots \: of \: {x}^{2} + 20x - 2020 = 0}$

We know

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm \implies\:ab = \dfrac{ - 2020}{1} = - 2020$

And

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm \implies\:a + b = - \dfrac{20}{1} = - 20$

Also, given that

$\red{\rm :\longmapsto\:c,d \: are \: the \: roots \: of \: {x}^{2} - 20x + 2020 = 0}$

$\rm \implies\:c + d = - \dfrac{( - 20)}{1} = 20$

and

$\rm \implies\:cd = \dfrac{2020}{1} = 2020$

Now, Consider

$\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d)$

$\sf \: = {ca}^{2} - {ac}^{2} + {da}^{2} - {ad}^{2} + {cb}^{2} - {bc}^{2} + {db}^{2} - {bd}^{2}$

$\sf \: = {a}^{2}(c + d) + {b}^{2}(c + d) - {c}^{2}(a + b) - {d}^{2}(a + b)$

$\sf \: = (c + d)( {a}^{2} + {b}^{2}) - (a + b)( {c}^{2} + {d}^{2})$

$\sf \: = 20( {a}^{2} + {b}^{2}) + 20( {c}^{2} + {d}^{2})$

$\sf \: = 20\bigg[ {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}\bigg]$

We know,

$\boxed{\tt{ { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta) }^{2} - 2 \alpha \beta \: }}$

So, using this, we get

$\sf \: = 20\bigg[ {(a + b)}^{2} - 2ab + {(c + d)}^{2} - 2cd\bigg]$

$\sf \: = 20\bigg[ {( - 20)}^{2} + 2(2020) + {(20)}^{2} - 2(2020)\bigg]$

$\sf \: = 20\bigg[ 400 + 400\bigg]$

$\sf \: = 20\bigg[ 800\bigg]$

$\sf \: = 16000$

Hence,

$\boxed{\tt{ \sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) = 16000}}$

<em>So, option (d) is correct.</em>

4 0

2 years ago