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Tems11 [23]
3 years ago
14

The product of two consecutive positive integers is 11 more than their sum. find integers

Mathematics
2 answers:
WITCHER [35]3 years ago
5 0

We will call N the first consecutive positive integer.

Then

N(N + 1) = N + (N +1) +11

N² +N = N + N +1 +11 = N² +N = 2N +12

N² + N - 2N -12 = 0

N² - N - 12 = 0

N = \frac{1+-\sqrt{1^{2} + 4*1*12}}{2*1}   = \frac{1*-\sqrt{49}}{2}  = \frac{1+-7}{2}

We only need positive integers, then

then N = (1 + 7)/2 = 8/2 = 4

and N + 1 = 4 + 1 = 5

ANSWER two positive consecutive integers are 4 and 5

checking

4*5 = 4+5 +11

20 = 9 +11 = 20 match!!

\textbf{Spymore}​

vodomira [7]3 years ago
3 0

n(n+1)=n+n+1+11\\n^2+n=2n+12\\n^2-n-12=0\\n^2+3n-4n-12=0\\n(n+3)-4(n+3)=0\\(n-4)(n+3)=0\\n=4 \vee n=-3\\\\-3\not>0\\n=4\\n+1=5

So, it's 4 and 5.

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One as it intersects only on the Y axis 
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3 years ago
Write two different rational functions whose graphs have the same end behaviour as the graph of y=3x^2
baherus [9]

Answer:

               y=x^2+5x+20\\ \\ y=8x^2+35

Explanation:

The <em>end behavior</em> of a <em>rational function</em> is the limit of the function as x approaches negative infinity and infinity.

Note that the the values of even functions are the same for ± x. That implies that their limits for ± ∞ are equal.

The limits of the quadratic function of general form y=ax^2+bx+c as x approaches negative infinity or infinity, when a  is positive, are infinity.

That is because as the absolute value of x gets bigger y becomes bigger too.

In mathematical symbols, that is:

\lim_{x \to -\infty}3x^2=\infty\\ \\ \lim_{x \to \infty}3x^2=\infty

Hence, the graphs of any quadratic function with positive coefficient of the quadratic term will have the same end behavior as the graph of y = 3x².

Two examples are:

         y=x^2+5x+20\\ \\ y=8x^2+35

5 0
4 years ago
Find the volume of the cylinder. height- 10 in. base- 3 in.
uysha [10]

The volume of the cylinder would be:

282.74

7 0
3 years ago
Determine the measure of each segment then indicate whether the statements are true or false
kupik [55]

Answer:

d_{AB}\ne d_{JK}

d_{AB}\ne \:d_{GH}

d_{GH}\ne \:d_{JK}

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

Step-by-step explanation:

Considering the graph

Given the vertices of the segment AB

  • A(-4, 4)
  • B(2, 5)

Finding the length of AB using the formula

d_{AB}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}

        =\sqrt{\left(2-\left(-4\right)\right)^2+\left(5-4\right)^2}

         =\sqrt{\left(2+4\right)^2+\left(5-4\right)^2}

         =\sqrt{6^2+1}

         =\sqrt{36+1}

        =\sqrt{37}

d_{AB}\:=\sqrt{37}

d_{AB}=6.08 units        

Given the vertices of the segment JK

  • J(2, 2)
  • K(7, 2)

From the graph, it is clear that the length of JK = 5 units

so

d_{JK}=5 units

Given the vertices of the segment GH

  • G(-5, -2)
  • H(-2, -2)

Finding the length of GH using the formula

d_{GH}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}

         =\sqrt{\left(-2-\left(-5\right)\right)^2+\left(-2-\left(-2\right)\right)^2}

          =\sqrt{\left(5-2\right)^2+\left(2-2\right)^2}

          =\sqrt{3^2+0}

           =\sqrt{3^2}

\mathrm{Apply\:radical\:rule\:}\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0

d_{GH}\:=\:3 units

Thus, from the calculations, it is clear that:

d_{AB}=6.08  

d_{JK}=5

d_{GH}\:=\:3

Thus,

d_{AB}\ne d_{JK}

d_{AB}\ne \:d_{GH}

d_{GH}\ne \:d_{JK}

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

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Step-by-step explanation:

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