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Sergeu [11.5K]
2 years ago
7

Find x: 3/(x-4)(x-7) + 6/(x-7)(x-13) + 15/(x-13)(x-28) - 1/x-28 = -1/20

Mathematics
1 answer:
Novay_Z [31]2 years ago
5 0

The value of x<em> </em>in the polynomial fraction 3/((x-4)•(x-7)) + 6/((x-7)•(x-13)) + 15/((x-13)•(x-28)) - 1/(x-28) = -1/20 is <em>x </em>= 24

<h3>How can the polynomial with fractions be simplified to find<em> </em><em>x</em>?</h3>

The given equation is presented as follows;

\frac{3}{(x - 4) \cdot (x - 7) }  + \frac{6}{(x - 7) \cdot (x - 13)   }  + +\frac{15}{(x - 13) \cdot (x - 28) } - \frac{1}{(x - 28)  } =  -  \frac{1}{20}

Factoring the common denominator, we have;

\frac{3\cdot(x - 13) \cdot(x - 28) + 6 \cdot(x - 4) \cdot(x - 28)  + 15 \cdot(x - 4) \cdot(x - 7)  - (x - 4) \cdot (x - 7)\cdot(x - 13)}{(x - 4) \cdot (x - 7)\cdot(x - 13) \cdot(x - 28)}   + =  -  \frac{1}{20}

Simplifying the numerator of the right hand side using a graphing calculator, we get;

By expanding and collecting, the terms of the numerator gives;

-(x³ - 48•x + 651•x - 2548)

Given that the terms of the numerator have several factors in common, we get;

-(x³ - 48•x + 651•x - 2548) = -(x-7)•(x-28)•(x-13)

Which gives;

\frac{-(x - 7) \cdot(x - 28)\cdot (x - 13)}{(x - 4) \cdot (x - 7)\cdot(x - 13) \cdot(x - 28)}   + =  -  \frac{1}{20}

Which gives;

\frac{-1}{(x - 4)}   + =  -  \frac{1}{20}

x - 4 = 20

Therefore;

  • x = 20 + 4 = 24

Learn more about polynomials with fractions here:

brainly.com/question/12262414

#SPJ1

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Write the perimeter of the triangle as a simplified polynomial. Then factor the polynomial
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c

Step-by-step explanation:

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We want to determine if the probability that a student enrolled in an accelerated math pathway is independent of whether the stu
Molodets [167]

Answer:

The correct option is (c).

Step-by-step explanation:

The complete question is:

The data for the student enrollment at a college in Southern California is:

                    Traditional          Accelerated            Total

                  Math-pathway     Math-pathway

Female              1244                       116                   1360

Male                  1054                       54                    1108

Total                  2298                     170                  2468

We want to determine if the probability that a student enrolled in an accelerated math pathway is independent of whether the student is female. Which of the following pairs of probabilities is not a useful comparison?

a. 1360/2468 and 116/170

b. 170/2468 and 116/1360

c. 1360/2468 and 170/2468

Solution:

If two events <em>A</em> and <em>B</em> are independent then:

P(A|B)=P(A)\\\\\&\\\\P(B|A)=P(B)

In this case we need to determine whether a student enrolled in an accelerated math pathway is independent of the student being a female.

Consider the following probabilities:

P (F|A) = \farc{116}{170}\\\\P(A|F)=\frac{116}{1360}\\\\P(A)=\frac{170}{2468}\\\\P(F)=\frac{1360}{2468}

If the two events are independent then:

P (F|A) = P(F)

&

P (A|F) = P (A)

But what would not be a valid comparison is:

P (A) = P(F)

Thus, the correct option is (c).

4 0
3 years ago
Can you help? Having a hard time figuring this out. Thanks
Dmitrij [34]

Answer:

3a) The value of  x = 56

3b) The measure of ∠ H T M = 90°

3c) The radius of the circle = 53

Step-by-step explanation:

3a) ∵ A F is a tangent to the circle O at point F

∵ Secant AH intersects circle O at point T

∴ (A F)² = (A T)(A H)

∴ 7( x + 7) = (21)² ⇒ ÷ 7

∴ x + 7 = 63

∴ x = 63 - 7 = 56

3b) ∵ HM is a diameter

∴ The measure of the arc HM = 180° ⇒ semi-circle

∵ ∠ H T M is inscribed angle subtended by the arc HM

∴ m ∠ H T M = half the measure of arc HM

∴ m ∠ H T M = 180° ÷ 2 = 90°

3c) ∵ Δ H T M is a right angle triangle at T

∴ (H M)² = (M T)² + (H T)² ⇒ Pythagorean theorem

∴ (H M)² = (90)² + (56)²

∴ (H M)² = 11236

∴ HM = \sqrt{11236} = 106

∴ OM = 106 ÷ 2 = 53

∵ OM is the radius of the circle O

∴ The radius  = 53

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g(x) <=-1

the output is less than or equal to -1

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