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lubasha [3.4K]
3 years ago
12

What is the simplified form of the quantity 4 x squared minus 25 over the quantity 2x minus 5 ?

Mathematics
2 answers:
Alenkasestr [34]3 years ago
7 0

Answer:

  B.  2x + 5, with the restriction x ≠ five over 2

Step-by-step explanation:

The given expression is the difference of squares, so factors as ...

  the product of the quantity 2x plus 5 and the quantity 2x minus 5 over the quantity 2x minus 5

You will note that the numerator and denominator have a common factor:

  the quantity 2x minus 5

Factoring that out gives ...

  2x + 5, with the restriction x ≠ five over 2 (x is restricted from being a value that makes the denominator zero.)

_____

<em>Comment on the form of the answer</em>

Since you have written your math expressions using words instead of symbols, we assume you can read them more easily that way. So, we have provided the explanation in the form you can most easily understand. (Personally, I prefer math symbols. They are more compact and tend to be less ambiguous.)

navik [9.2K]3 years ago
5 0

Answer:

B) 2x + 5, with the restriction x ≠ five over 2

Step-by-step explanation:

The given verbal expression is " 4 x squared minus 25 over the quantity 2x minus 5."

Now let's convert this to algebraic expression, we get

\frac{4x^2 - 25}{2x - 5}

Now let's factorize the numerator.

4x^2 - 25\\= (2x)^2 - 5^2\\= (2x -5)(2x + 5)\\

We used the identity (a^2 - b^2) = (a -b)(a+b)

= \frac{(2x - 5)(2x +5)}{2x - 5}

Here we have (2x - 5) both in the numerator and in the denominator, we can cancel out.

= (2x + 5) where x ≠ 5/2

So the answer is B) 2x + 5, with the restriction x ≠ five over 2

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If it’s costs 60 cents for 125 grammes. How much would 100 grammes cost?
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Multiply this by 100 to get the answer:

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4 years ago
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The concentration of particles in a suspension is 50 per mL. A 5 mL volume of the suspension is withdrawn. a. What is the probab
kolezko [41]

Answer:

(a) 0.6579

(b) 0.2961

(c) 0.3108

(d) 240

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of particles in a suspension.

The concentration of particles in a suspension is 50 per ml.

Then in 5 mL volume of the suspension the number of particles will be,

5 × 50 = 250.

The random variable <em>X</em> thus follows a Poisson distribution with parameter, <em>λ</em> = 250.

The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 10.  

The mean of the approximated distribution of X is:

μ = λ  = 250

The standard deviation of the approximated distribution of X is:

σ = √λ  = √250 = 15.8114

Thus, X\sim N(250, 250)

(a)

Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:

P(235

                             =P(-0.95

Thus, the value of P (235 < <em>X</em> < 265) = 0.6579.

(b)

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

                             =P(-0.28

Thus, the value of P(48.

(c)

A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

                             =P(-0.40

Thus, the value of P(48.

(d)

Let the sample size be <em>n</em>.

P(48

                             0.95=P(-z

The value of <em>z</em> for this probability is,

<em>z</em> = 1.96

Compute the value of <em>n</em> as follows:

z=\frac{\bar X-\mu}{\sigma/\sqrt{n}}\\\\1.96=\frac{48-50}{15.8114/\sqrt{n}}\\\\n=[\frac{1.96\times 15.8114}{48-50}]^{2}\\\\n=240.1004\\\\n\approx 241

Thus, the sample selected must be of size 240.

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Answer:

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hhcxxghjo9

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