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Citrus2011 [14]
2 years ago
6

Factor completely 2x2 – X – 4.

Mathematics
1 answer:
goblinko [34]2 years ago
8 0

2x {}^{2}  - x - 4

x =  \frac{ - ( - 1) +  \sqrt{1 - 4(2)( - 4)} }{2(2)}  =  \frac{1  +  \sqrt{33} }{4}

x =  \frac{1 -  \sqrt{33} }{4}

(x -  \frac{1 +  \sqrt{33} }{4} )(x -  \frac{1 -  \sqrt{33} }{4} )

(4x - 1  -   \sqrt{33} )(4x - 1 +  \sqrt{33} )

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Step-by-step explanation:


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3 years ago
Evaluate the following. Enter the answer as a simplified fraction.<br> sin(-11pi/6)cos(4pi/3)
Stels [109]

Answer:  -25/100 or -1/4

Sorry if its wrong, Happy To Help

Step-by-step explanation:

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3 years ago
If a translation of T-3,-8 (x,y) is applied to square ABCD, what is the y-coordinate of B'?
yarga [219]

Answer:

  -6

Step-by-step explanation:

The y-coordinate of B is 2. Adding -8 to it makes the y-coordinate of B' be -6.

  2 -8 = -6

3 0
3 years ago
Read 2 more answers
Since at t=0, n(t)=n0, and at t=[infinity], n(t)=0, there must be some time between zero and infinity at which exactly half of t
Ostrovityanka [42]

An expression for this timet-half will be given as ln(2) / λ ≈ 0.693 / λ

<h3>What is an expression?</h3>

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication and division.

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.

1) Equation given:

← 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:

N9t) = Ne^{-\lambda t}

3) Solving for α (remember α is λ)

N_{t-half}=\dfrac{N_o}{2}=N_oe^{-\alpha t}

\dfrac{1}{2} = e^{-\alpha t}2=e^{-alpha t}\alpha t=In(2)

αt ≈ 0.693

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

Therefore expression for this timet-half will be given as ln(2) / λ ≈ 0.693 / λ

To know more about an expression follow

brainly.com/question/723406

#SPJ4

3 0
2 years ago
There are 130 people in a
Sonja [21]

Answer:

If 11 people use all three facilities, this means:

22 − 11 or 11 are using the gym and the pool but no the track,

29 − 11 or 18 are using the pool and the track but no the gym,

25 − 11 or 14 are using the gym and the track but no the pool.

How many people use the gym only?

130 − (14 + 11 + 11) = 94

How many people use the pool only?

62 − (18 + 11 + 11) = 22

How many people use the track only?

58 − (14 + 11 + 18) = 15

How many people go in for sports?

94 + 22 + 15 + 11 + 14 + 18 + 11 = 185

How many people use exactly one of the facilities?

94 + 22 + 15 = 131

What is the probability that the person uses exactly one of the facilities?

= 131/185 ≈ 0.7081

Step-by-step explanation:

its a bit confusing

Hope It Helps !!!!

7 0
2 years ago
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