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xenn [34]
3 years ago
15

x and x+1 are the two continuous natural numbers, then (i) Write the next two natural numbers. (ii) Find the product of first an

d the fourth number. (iii) Find the product of second and the third number. (iv) Find the relation between these products.
Mathematics
2 answers:
Gala2k [10]3 years ago
7 0

Answer:

(i)  X+2 and x+3;

(ii) x( x+4)= X^2 +4x;

(iii) x( x+4)=a

    (x+1)(x+3)= a+3  

Step-by-step explanation:

(i)  X+2 and x+3;

(ii) x( x+4)= X^2 +4x;

(iii) (x+1)(x+3)=x^2+3X+X+3=x^2+4x+3;

x(x+4)=a

(x+1)(x+3)= a+3

Alinara [238K]3 years ago
3 0

Step-by-step explanation:

x is followed by x+1 so each time we add one

our sequence goes like this:

  • x
  • x+1
  • x+1+1⇒ x+2
  • x+2+1⇒x+3

the first number is x and the fourth one is x+3

  • x*(x+3)
  • x²+3x

the second number is x+1 and the third one is x+2

  • (x+1)(x+2)
  • x²+2x+x+2
  • x²+3x+2

the first product is x²+3x and the second one is x²+3x+2

Let P be x²+3x and P' be x²+3x+2

  • P'-P = 2
  • P' = P+2
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8 divide by 1 1/4 KCF
Diano4ka-milaya [45]

Answer:

32/5

Step-by-step explanation:

K=Keep the first Fraction (You can rewrite 8 as 8/1 for when you multiply across.)

C=Change the Division Sign to a Multiplication Sign

F=Flip the Second Fraction

First, rewrite 1 1/4 as 5/4 (A mixed number can be changed to a fraction by multiplying the outside whole number by the denominator or bottom number of the fraction, then add the new rewritten whole number and the original fractional piece. In this case you would multiply the outside 1 by the bottom 4 and add it to the original 1/4. 4/4+1/4=5/4)

Second, Flip 5/4 to 4/5 and change your equation so it now reads: 8/1 x 4/5

Third, use simple fraction multiplication and multiply across to get 8/1 x 4/5=32/5

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3 years ago
Fine the equation of the line: slope = -2 y-intercept= 5
soldi70 [24.7K]
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3 0
2 years ago
Qhat is 3 groupps of 1/8
iragen [17]

Answer: 3/8ths

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4 0
2 years ago
Y''+y'+y=0, y(0)=1, y'(0)=0
mars1129 [50]

Answer:

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form ay''+by'+cy=0.

Given: y''+y'+y=0

Let y=e^{rt} be it's solution.

We get,

\left ( r^2+r+1 \right )e^{rt}=0

Since e^{rt}\neq 0, r^2+r+1=0

{ we know that for equation ax^2+bx+c=0, roots are of form x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} }

We get,

y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}

For two complex roots r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta, the general solution is of form y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )

i.e y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Applying conditions y(0)=1 on e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right ), c_1=1

So, equation becomes y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

On differentiating with respect to t, we get

y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )

Applying condition: y'(0)=0, we get 0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}

Therefore,

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

3 0
3 years ago
Someone help me answer this and show work please!
timurjin [86]

Answer:

Simple

Rise/Run

for one week they run 10 miles

2 weeks 20 miles

3 weeks 30 miles

4 weeks 40 miles

the rate of change would be 10/1 (Why? Because you rise +10 each time per week.)

divide that it would be 10 (:

gOoD lUCk

second answer copied me so.

7 0
3 years ago
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