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dsp73
3 years ago
15

Which term can be used in the blank of 36x3−22x2−__ so the greatest common factor of the resulting polynomial is 2x? Select two

options. 2 4xy 12x 24 44y
Mathematics
2 answers:
maria [59]3 years ago
3 0

Answer:

4xy and 12x

Step-by-step explanation:

In this question, we are to select which of the options to put into the blank space that will make the greatest common factor of the polynomial 2x. We are to select two options

Now, looking at the blank space, we should understand that , any answer we shall be choosing shall be an answer that contains x alongside the number.

This makes 24 and 44y out of the way.

Now let’s look at 4xy and 12x. Let’s insert them into the blank space;

36x^3-22x^2 - 4xy which gives : 2x(18x^2-11x-2y)

and 36x^3-22x^2-12x = 2x(18x^2-11x-6)

The reason why we will not use 44y and 24 is because 2x is not a factor of either.

alexandr402 [8]3 years ago
3 0

Answer:

Answer:4xy and 12x

Step-by-step explanation:

The could both be a product of 2x

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iogann1982 [59]
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The answer to you're question is C, because squared means to the power of two. So, 39 * 39 is 1,521.

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3 years ago
Which of the following steps could be used to solve the equation 93 = x + 58? A. Add 58 to both sides. B. Subtract 58 from both
mario62 [17]

Answer:

Subtract 58 from both sides

Step-by-step explanation:

93 = x + 58

Subtract 58 from both sides

93- 58 = x+58-58

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7 0
3 years ago
Given logx 2=p and logx 7=q , express log7 4x² in terms of p and q​
kodGreya [7K]

Answer:

[2(p + 1)]/q

Step-by-step explanation:

logx 2 = p

logx 7 = q

log7 4x² = log7 (2x)²

= (logx (2x)²)/(logx 7)

= (2 logx 2x)/(logx 7)

= (2 logx 2 + 2 logx x)/(logx 7)

= (2p + 2)/q

= [2(p + 1)]/q

5 0
3 years ago
Since at t=0, n(t)=n0, and at t=∞, n(t)=0, there must be some time between zero and infinity at which exactly half of the origin
Airida [17]
Answer: t-half = ln(2) / λ ≈ 0.693 / λ

Explanation:

The question is incomplete, so I did some research and found the complete question in internet.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=No e−λt,

where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

Answer:

1) Equation given:

N(t)=N _{0} e^{-  \alpha  t} ← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) =  No / 2

So, replace in the given equation:

N_{t-half} =  N_{0} /2 =  N_{0}  e^{- \alpha t}

3) Solving for α (remember α is λ)

\frac{1}{2} =  e^{- \alpha t} 

2 =   e^{ \alpha t} 

 \alpha t = ln(2)

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ




4 0
3 years ago
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