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yulyashka [42]
2 years ago
7

Which one is NOT a monomial

Mathematics
2 answers:
lesantik [10]2 years ago
7 0
<h3>Answer: Choice A</h3>

This is because 3x-4 is a binomial consisting of the two terms 3x and -4

A monomial has only one term, which choices B through D reflect that.

Valentin [98]2 years ago
3 0
A because it’s a binomial
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if your answer is "No," how much more lacquer does she need? if your answer is "yes," how much lacquer will she have left over?
babunello [35]

my answer is no ,yes ,no

So ,maybe u find the answer

Step-by-step explanation:

if your answer is "No," how much more lacquer does she need? if your answer is "yes," how much lacquer will she have left over?AND if it is also no ,yes , no...show your work to justify your answer.

3 0
3 years ago
Help, please<br> A: 2<br> B: -2<br> C: 1/2<br> D:-1/2
Anna [14]

Answer:

\left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \sqrt{x} \left \{ {{y=2} \atop {x=2}} \right. \frac{x}{y} \alpha \alpha \alpha =\beta

Step-by-step explanation:

4 0
3 years ago
Given a = -3, b = 5, and c = -8, then a 2 b + c is _____.<br><br> -38<br> -53<br> 37<br> 22
AnnyKZ [126]
This is how you would arrive to the answer 37

4 0
3 years ago
Read 2 more answers
I’m lost again can somebody help asap ??
serious [3.7K]
Fully detailed and equipped information process right below. The answer for this question is "25" (simplified form).

\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(5^{- 5} \times 2^8 \times 1 \Bigg)^{- 2}}

\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(2^8 \times 1 \times \dfrac{1}{5^5} \Bigg)^{- 2}}

\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(1 \times \dfrac{2^8 \times 1}{5^5} \Bigg)^{- 2}}

\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(1 \times \dfrac{256}{3125} \Bigg)^{- 2}}

\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(\dfrac{3125}{256} \Bigg)^2}

\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \dfrac{3125^2}{256^2}}

\mathbf{2^{28} \times \Bigg(\dfrac{1}{2^3 \times 5^2} \Bigg)^4 \times \dfrac{3125^2}{256^2}}

\mathbf{2^{28} \times \Bigg(\dfrac{1^4}{(2^3 \times 5^2)^4} \Bigg) \times \dfrac{3125^2}{256^2}}

\mathbf{2^{28} \times \Bigg(\dfrac{1^4}{(5^2)^4 \times (2^3)^4} \Bigg) \times \dfrac{3125^2}{256^2}}

\mathbf{2^{28} \times \dfrac{1}{2^{12} \times 5^8} \times \dfrac{3125^2}{256^2}}

\mathbf{\dfrac{3125^2 \times 1 \times 2^{28}}{256^2 \times 5^8 \times 2^{12}}}

\mathbf{\dfrac{2^{(28 - 12)} \times 3125^2}{256^2 \times 5^8}}

\mathbf{\dfrac{2^{16} \times 3125^2}{256^2 \times 5^8}}

\mathbf{\dfrac{2^{16} \times (5^5)^2}{(2^8)^2 \times 5^8}}

\mathbf{\dfrac{2^{16} \times 5^{10}}{2^{16} \times 5^8}}

\mathbf{\dfrac{2^{16} \times 5^{10 - 8}}{2^{16}}}

\mathbf{\dfrac{5^{10 - 8}}{1}}

\mathbf{\therefore \quad 5^2}

\mathbf{\therefore \quad 25}

\boxed{\mathbf{\underline{\therefore \quad Final \: \: Answer \: \: is: \: 25}}}

Hope it helps.
7 0
3 years ago
What scenario could be modeled by the graph below?
Dafna1 [17]

Answer:

I would believe that it is A.

Step-by-step explanation:

The fact that A has 5, and x is minus two times, 5/2= 2.5, and that is where the second arrowhead is pointing to on the x axis.

6 0
3 years ago
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