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LUCKY_DIMON [66]
3 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)​
Mathematics
1 answer:
Paha777 [63]3 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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Can someone help with this?
Bumek [7]
4.5/3 = x/5

then:

x = 5 * 4.5 / 3 = 15/2 = 7.5


hope that helps

6 0
3 years ago
Choose the product of –3y2(–4y3 + y – 9).
Nina [5.8K]

Answer:

D

-3 * (-4) matches, as well as the powers (I guess you wanted to communicate that these are powers, but it would also be correct if these where factors)

6 0
3 years ago
Simplify the fraction.<br><br><br> what is 42 over 98
Darya [45]
42/98=3/7
Have a great day :)
4 0
3 years ago
Read 2 more answers
Which piece of additional information can be used to prove △CEA ~ △CDB?
Slav-nsk [51]

<u>Answer-</u>

<em>The correct answer is</em>

<em>∠BDC and ∠AED are right angles</em>

<u>Solution-</u>

In the ΔCEA and ΔCDB,

m\angle BCD=m\angle ACE

As this common to both of the triangle.

If ∠BDC and ∠AED are right angles, then m\angle E=90=m\angle D

Now as

∠BCD = ∠ACE and ∠BDC = ∠AED,

∠DBC and ∠EAC will be same. (as sum of 3 angles in a triangle is 180°)

Then, ΔCEA ≈ ΔCDB

Therefore, additional information can be used to prove ΔCEA ≈ ΔCDB is ∠BDC and ∠AED are right angles.


5 0
3 years ago
Read 2 more answers
If cos(θ)=2853 with θin Q IV, what is sin(θ)?
marysya [2.9K]

Answer: \sin \theta=\frac{-45}{53}

Step-by-step explanation:

Since we have given that

\cos\theta=\frac{28}{53}

And we know that θ is in the Fourth Quadrant.

So, Except cosθ and sec θ, all trigonometric ratios will be negative.

As we know the "Trigonometric Identity":

\cos^2\theta+\sin^2\theta=1\\\\\sin \theta=\sqrt{1-\cos^2\theta}\\\\\sin \theta=\sqrt{1-(\frac{28}{53})^2}=\sqrt{\frac{53^2-28^2}{53^2}}\\\\\sin \theta=\sqrt{\frac{2025}{53^2}}\\\\\sin \theta=\frac{45}{53}

It must be negative due to its presence in Fourth quadrant.

Hence, \sin \theta=\frac{-45}{53}

7 0
3 years ago
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