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3 years ago

Find a pair of square numbers which give a difference of 39

Mathematics

1 answer:

nadya68 [22]3 years ago

5 0

64-25=39 All the square numbers are 1,4,9,16,25,36,49,64,81,100

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Drawing 1/5 the original with a ship 14 inches

Shalnov [3]

The scaled down size is 1/5 time 14
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3 0

3 years ago

Show work and explain with formulas.

vova2212 [387]

Answer:

$\large\boxed{4.\ S_{13}=260}$

$\large\boxed{5.\ \sum\limits_{k=1}^7(2k+5)=91}$

$\large\boxed{6.\ a_1=17,\ a_2=26,\ a_3=35}$

Step-by-step explanation:

We have:

$a_6=18,\ a_{13}=32$

These are the terms of the arithmetic sequence.

We know:

$a_n=a_1+(n-1)d$

Therefore

$a_6=a_1+(6-1)d\ \text{and}\ a_{13}=a_1+(13-1)d\\\\a_6=a_1+5d\ \text{and}\ a_{13}=a_1+12d\\\\a_{13}-a_6=(a_1+12d)-(a_1+5d)\\\\a_{13}-a_6=a_1+12d-a_1-5d\\\\a_{13}-a_6=7d$

Substitute a₆ = 18 and a₁₃ = 32:

$7d=32-18$

$7d=14$ divide both sides by 7

$d=2$

$a_6=a_1+5d$

$18=a_1+5(2)$

$18=a_1+10$ subtract 10 from both sides

$8=a_1\to a_1=8$

The formula of a sum of terms of an arithmetic sequence:

$S_n=\dfrac{a_1+n_1}{2}\cdot n$

Substitute a₁ = 8, a₁₃ = 32 and n = 13:

$S_{13}=\dfrac{8+32}{2}\cdot13=\dfrac{40}{2}\cdot13=20\cdot13=260$

===========================================

We have

$\sum\limits_{k=3}^7(2k+5)\to a_k=2k+5$

Calculate $a_{k+1}$

$a_{k+1}=2(k+1)+5=2k+2+5=2k+7$

Calculate the difference:

$a_{k+1}-a_k=(2k+7)-(2k+5)=2k+7-2k-5=2$

It's the arithmetic sequence with first term

$a_1=2(1)+5=2+5=7$

and common difference d = 2.

The formula of a sum of terms of an arithmetic sequence:

$S_n=\dfrac{2a_1+(n-1)d}{2}\cdot n$

Substitute n = 7, a₁ = 7 and d = 2:

$S_7=\dfrac{2(7)+(7-1)(2)}{2}\cdot7=\dfrac{14+(6)(2)}{2}\cdot7=\dfrac{14+12}{2}\cdot7=\dfrac{26}{2}\cdot7=(13)(7)\\\\S_7=91$

===========================================

We have:

$a_1=17,\ a_n=197,\ S_n=2247$

The formula for the n-th term of an arithmetic sequence:

$a_n=a_1+(n-1)d$

The formula of the sum of terms of an arithmetic sequence:

$S_n=\dfrac{2a_1+(n-1)d}{2}\cdot n$

Substitute:

$(1)\qquad 197=17+(n-1)d\\\\(2)\qquad2247=\dfrac{(2)(17)+(n-1)d}{2}\cdot n$

Convert the first equation:

$197=17+(n-1)d$ subtract 17 from both sides

$180=(n-1)d$ Substitute it to the second equation:

$2247=\dfrac{34+180}{2}\cdot n\\\\2247=\dfrac{214}{2}\cdot n$

$2247=107n$ divde both sides by 107

$21=n\to n=21$

Put the value of n to the equation (n - 1)d = 180:

$(21-1)d=180$

$20d=180$ divide both sides by 20

$d=9$

Therefore we have the explicit formula for the nth term of an arithmetic sequence:

$a_n=17+(n-1)(9)\\\\a_n=17+9n-9\\\\a_n=9n+8$

Put n = 1, n = 2 and n = 3:

$a_1=9(1)+8=9+8=17\qquad\text{CORRECT :)}\\\\a_2=9(2)+8=18+8=26\\\\a_3=9(3)+8=27+8=35$

4 0

3 years ago

Aliyah bought a car for $32,000. It is now worth $5000 less than half of what she purchased it for. To the nearest tenth of a pe

Reil [10]

32,000 x 0.5 = 16000

16000- 5000 = 11000

11000 / 32000= 0.34375

1- 0.344= 0.656

Answer : D

3 0

3 years ago

When two objects reach the same time-space coordinates, they generally collide with each other. How is this different from what

nikklg [1K]

Answer:

Waves superimpose upon each other when they collide, while objects do not

Step-by-step explanation:

The main difference between the collision of waves and the collision of objects is simply the superposition principle.

When waves collide, they do not do so in the same way objects do. The superposition principle explains that waves can either collide in a constructive or destructive manner.

Case A: Waves colliding in a constructive manner

When waves collide in a constructive manner, this means that they are in phase, in simpler terms, it means that they have the same shape as they move through space-time. Constructive collision leads to a formation of a bigger wave with a higher amplitude. This is how stereo speakers operate. They produce louder sounds by releasing the same audio waves, causing them to superimpose upon each other.

Case B: Waves colliding in a destructive manner:

When waves are out of phase(i.e do not have the same shape as they move through space-time) and they collide, they try to cancel each other out, leading to a new wave with a weaker amplitude. This is how noise-cancelling headphones work. They emit an equal and opposite wave sound to the noise around your ears, thus cancelling it out.

4 0

3 years ago

What is the difference? StartFraction 2 x + 5 Over x squared minus 3 x EndFraction minus StartFraction 3 x + 5 Over x cubed minu

Sunny_sXe [5.5K]

Answer:

First option is the correct option.

Step-by-step explanation:

$\frac{2x + 5}{ {x}^{2} - 3x } - \frac{3x + 5}{ {x}^{3} - 9x } - \frac{x + 1}{ {x}^{2} - 9 }$

Factor out X from the expression

$\frac{2x + 5}{x(x - 3)} - \frac{3x + 5}{x( {x}^{2} - 9)} - \frac{x + 1}{ {x}^{2} - 9}$

Using ${a}^{2} - {b}^{2} = (a - b)(a + b)$ , factor the expression

$\frac{2x + 5}{x(x - 3)} - \frac{3x + 5}{x(x - 3)(x + 3) } - \frac{x + 1}{(x - 3)(x + 3)}$

Write all numerators above the Least Common Denominators x ( x - 3 ) ( x + 3 )

$\frac{(x + 3) \times (2x - 5) - (3x + 5) - x \times (x + 1)}{x(x - 3)(x + 3)}$

Multiply the parentheses

$\frac{2 {x}^{2} + 5x + 6x + 15 - (3x + 5) - x(x + 1)}{x(x - 3)(x + 3)}$

When there is a (-) in front of an expression in parentheses, change the sign of each term in the expression