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photoshop1234 [79]
2 years ago
5

If p is a positive integer, and 4 is the remainder when p-8 is divided by 5, which of the following

Mathematics
1 answer:
malfutka [58]2 years ago
6 0
<h2>sorry I can't help you to solve the equation</h2>
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Please explain how they plugged the removable discontinuity in this question.
kondor19780726 [428]

In the limit

\displaystyle \lim_{x\to c} f(x)

we're interested in the value that f(x) converges to as x gets closer to c. So in fact x\neq c.

In the given example, f(x) is factorized to reveal a common factor of x-1 in the numerator and denominator. We have x\neq1 if x\to1, so x-1\neq0 so we can simplify

\dfrac{x-1}{x-1} = 1

and *remove* the discontinuity.

Then

\displaystyle \lim_{x\to1} \frac{2(x-1)}{(x+1)(x-1)} = \lim_{x\to1} \frac2{x+1} = \frac2{1+1} = 1

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Which relation is not a function
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5 0
2 years ago
Plz help I have no idea how to answer it​
saveliy_v [14]

5. Answer: see explanation

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

If the roots are m and 3m, then x = m and x = 3m

⇒ x - m = 0    and      x - 3m = 0

⇒ (x - m)(x - 3m) = 0

⇒ x² - 4mx + 3m² = 0

Since x² + px + q = 0   then  p = -4m   and q = 3m²

3p² = 3(-4m)²               16q = 16(3m²)

      = 3(16m²)                      = 48m²

      = 48m²

3p² = 48m² = 16q     ⇒           3p² = 16q

**********************************************************************************************

6. Answer: 8 or 18

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

The Area of the entire rectangle (A = L × w) is 12 × 10 = 120

The Area of the shaded region is 72, so the Area of the non-shaded region is 120 - 72 = 48.

There are two non-shaded triangles.

  • Bigger non-shaded Δ: L = 12-x, w = 10  ⇒ A = \dfrac{10(12-x)}{2}
  • Smaller non-shaded Δ: L = x, w = x ⇒ A=\dfrac{x(x)}{2}

Combine the Areas of both triangles and set it equal to the Area of the non-shaded region:

\dfrac{x(x)}{2}+\dfrac{10(12-x)}{2}=48\\\\\\\dfrac{x^2}{2}+\dfrac{120-10x}{2}=48\\\\\\\dfrac{x^2-10x+120}{2}=48\\\\\\x^2-10x+120=96\\\\x^2-10x+24=0\\\\(x-4)(x-6)=0\\\\x - 4 = 0\quad or\quad x-6=0\\x=4\qquad or\qquad x=6\\\\\text{Based on the image provided, it appears that the x-value will be the}\\\text{smallest of the two possible solutions (4), but both solutions are valid.}

Area of ΔBEF:

\text{when x = 4, A =}\ \dfrac{4(4)}{2}=\dfrac{16}{2}=8\\\\\\\text{when x = 6, A =}\ \dfrac{6(6)}{2}=\dfrac{36}{2}=18

5 0
3 years ago
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