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zloy xaker [14]
3 years ago
9

Simplify (7x+42)/(x^2+13x+42)

Mathematics
2 answers:
Colt1911 [192]3 years ago
7 0
\frac{7x+42}{x^2+13x+42}=  \frac{7(x+6)}{x^2+7x+6x+42}=   \frac{7(x+6)}{x(x+7)+6(x+7)}=  \frac{7(x+6)}{(x+6)(x+7)}=  \frac{7}{x+7}
grandymaker [24]3 years ago
4 0
<h2>Answer:</h2>

The simplified expression is given by:

                \dfrac{7}{x+7}

<h2>Step-by-step explanation:</h2>

We are asked to simplify the expression which is given in terms of variable x as:

             =\dfrac{7x+42}{x^2+13x+42}

The given expression is a rational expression as it is a polynomial expression in the denominator as well as in the numerator.

Now we know that:

7x+42=7(x+6)

and

x^2+13x+42=x^2+7x+6x+42\\\\\\i.e.\\\\\\x^2+13x+42=x(x+7)+6(x+7)\\\\\\i.e.\\\\\\x^2+13x+42=(x+7)(x+6)

Hence, we get:

\dfrac{7x+42}{x^2+13x+42}=\dfrac{7(x+6)}{(x+6)(x+7)}\\\\\\i.e.\\\\\\\dfrac{7x+42}{x^2+13x+42}=\dfrac{7}{x+7}

            Hence, the simplified expression is:

                    \dfrac{7}{x+7}

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Evaluate the expression if k = 5 and w = 3 *<br> 3k + 2w - 4K
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Please help me with this.
JulijaS [17]

The <em>speed</em> intervals such that the mileage of the vehicle described is 20 miles per gallon or less are: v ∈ [10 mi/h, 20 mi/h] ∪ [50 mi/h, 75 mi/h]

<h3>How to determine the range of speed associate to desired gas mileages</h3>

In this question we have a <em>quadratic</em> function of the <em>gas</em> mileage (g), in miles per gallon, in terms of the <em>vehicle</em> speed (v), in miles per hour. Based on the information given in the statement we must solve for v the following <em>quadratic</em> function:

g = 10 + 0.7 · v - 0.01 · v²      (1)

An effective approach consists in using a <em>graphing</em> tool, in which a <em>horizontal</em> line (g = 20) is applied on the <em>maximum desired</em> mileage such that we can determine the <em>speed</em> intervals. The <em>speed</em> intervals such that the mileage of the vehicle is 20 miles per gallon or less are: v ∈ [10 mi/h, 20 mi/h] ∪ [50 mi/h, 75 mi/h].

To learn more on quadratic functions: brainly.com/question/5975436

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3 0
2 years ago
A doctor is measuring the mean systolic blood pressure of female students at a large college. Systolic blood pressure is known t
Nana76 [90]

Answer:

The correct option is (b).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

CI=\bar x\pm z_{\alpha/2}\times\frac{SD}{\sqrt{n}}

The confidence interval for population mean can be computed using either the <em>z</em>-interval or <em>t</em>-interval.

The <em>t</em>-interval is used if the following conditions are satisfied:

  • The population standard deviation is not known
  • The sample size is large enough
  • The population from which the sample is selected is normally distributed.

For computing a (1 - <em>α</em>)% confidence interval for population mean , it is necessary for the population to normally distributed if the sample selected is small, i.e.<em>n</em> < 30, because only then the sampling distribution of sample mean will be approximated by the normal distribution.

In this case the sample size is, <em>n</em> = 28 < 30.

Also it is provided that the systolic blood pressure is known to have a skewed distribution.

Since the sample is small and the population is not normally distributed, the  sampling distribution of sample mean will not be approximated by the normal distribution.

Thus, no conclusion can be drawn from the 90% confidence interval for the mean systolic blood pressure.

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5 0
3 years ago
PLEASE NEEED HEELPP
krok68 [10]
There are many systems of equation that will satisfy the requirement for Part A.
an example is y≤(1/4)x-3 and y≥(-1/2)x-6
y≥(-1/2)x-6 goes through the point (0,-6) and (-2, -5), the shaded area is above the line. all the points fall in the shaded area, but
y≤(1/4)x-3 goes through the points (0,-3) and (4,-2), the shaded area is below the line, only A and E are in the shaded area. 
only A and E satisfy both inequality, in the overlapping shaded area.

 
Part B. to verify, put the coordinates of A (-3,-4) and E(5,-4) in both inequalities to see if they will make the inequalities true. 
 for y≤(1/4)x-3: -4≤(1/4)(-3)-3
-4≤-3&3/4 This is valid.
For y≥(-1/2)x-6: -4≥(-1/2)(-3)-6
-4≥-4&1/3 this is valid as well. So Yes, A satisfies both inequalities. 
Do the same for point E (5,-4)

Part C: the line y<-2x+4 is a dotted line going through (0,4) and (-2,0)
the shaded area is below the line
farms A, B, and D are in this shaded area. 
8 0
3 years ago
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