If you will solve this math 50 point + brainlists= 5+6+6+83+15×53___

A positive integer is twice another.the sum of the reciprocal of the two positive integer is 3/14. Find the integers

icang [17]

Answer:

$\huge\boxed{14\ \text{and}\ 7}$

Step-by-step explanation:

$n,\ m-\text{positive integer}\\\\n=2m-\text{a positive integer is twice another}\\\\\dfrac{1}{n}+\dfrac{1}{m}=\dfrac{3}{14}-\text{the sum of the reciprocal of the two positive integer is }\ \dfrac{3}{14}\\\\\text{We have the system of equations:}\\\\\left\{\begin{array}{ccc}n=2m&(1)\\\dfrac{1}{n}+\dfrac{1}{m}=\dfrac{3}{14}&(2)\end{array}\right$

$\text{Substitute (1) to (2):}\\\\\dfrac{1}{2m}+\dfrac{1}{m}=\dfrac{3}{14}\\\\\dfrac{1}{2m}+\dfrac{1\cdot2}{m\cdot2}=\dfrac{3}{14}\\\\\dfrac{1}{2m}+\dfrac{2}{2m}=\dfrac{3}{14}\\\\\dfrac{1+2}{2m}=\dfrac{3}{14}\\\\\dfrac{3}{2m}=\dfrac{3}{14}\Rightarrow2m=14\qquad\text{divide both sides by 2}\\\\\dfrac{2m}{2}=\dfrac{14}{2}\\\\\boxed{m=7}$

$\text{Substitute it to (1):}\\\\n=2\cdot7\\\\\boxed{n=14}$

8 0

3 years ago

You buy a 1:1000 scale model of the Statue of Liberty during a trip to New York City. The height of the model is 9.3 centimeters

luda_lava [24]

Scale is 1:1000
Thus 9.3cm the actual is 9.3×1000= 9300cm = 93m

8 0

3 years ago

Tim explained a function in words and Paul wrote an equation. Tim The amount of money in Marshas savings account increases at a

soldier1979 [14.2K]

Answer:

Tim has the smaller y-intercept = 2780

Step-by-step explanation:

Tim explains that Marshas saving increases by 225 per month. After 8 months, Marsha's has 4580 in his account.

Let m be the gradient(change per month)

Let y be the initial money in Marsha's account.

Let y1 be the amount in Marsha's account after 8 months

y1 = 4580

Let x be the initial month d money was put into the account.

Let x1 be the number of months it takes for money to increase to 4580

x1= 8

(y - y1) /(x - x1) = m

(y - 4580)/ (x - 8) = 225

y - 4580 = 225(x - 8)

y - 4580 = 225x - 1800

y = 225x - 1800 + 4580

y = 225x + 2780

Comparing with y = mx + c, c(intercept) = 2780

For Paul,

y - 1400 = 56(x + 26)

y -1400 = 56x + 1456

y = 56x + 1456 + 1400

y = 56x + 2856

Comparing with y = mx + c, c(intercept) = 2850

Therefore comparing Tim's Y intercept and Paul's Y intercept,

2780 < 2856

Tim's function smaller y intercept = 2780

7 0

3 years ago

a piggy bank contained $14.55 in quarters dimes and nickels if there were three more than twice as many dimes as nickels and thr

GarryVolchara [31]

Let x=number of nickels in the piggy bank
(2x+3)=number of dimes in the piggy bank
(3x-3)=number of quarters in the piggy bank

.05x+.10(2x+3)+.25(3x-3)=14.55
.05x+.20x+.30+.75x-.75=14.55



1.00x=14.55+.45
x=15
2x+3=33
3x-3=42

The number of nickles in the piggy bank would be 15.

The number of dimes in the piggy bank would be 33

The number of quarters in the piggy bank would be 42. 

Hope I was able to help and that this was useful to you!!! :)

6 0

3 years ago

Read 2 more answers

If points A and B are randomly placed on the circumference of a circle with a radius of 2 cm, what is the probability that the l

OLga [1]

Answer:

The probability that the length of chord AB > 2 cm is 2/3 ⇒ D

Step-by-step explanation:

* Lets explain how to solve the problem

- Points A and B are randomly placed on the circumference of a

circle with a radius of 2 cm

- We need to find the probability that the length of chord AB is

greater than 2 cm

- Probability = required length of arc / circumference

- Assume that the length of the chord AB is 2 and the center of

the circle is labeled M

- The radius of the circle is 2 and AB = 2

- In Δ AMB

∵ MA= MB = 2 radii

∵ AB = 2

∴ Δ AMB is an equilateral triangle

∴ m∠AMB = π/3

∵ The length of an arc = r Ф , where r is the radius of the circle and

Ф is the central angle subtended by this arc

∵ r = 2 , Ф = π/3

∴ The length of arc AB = 2 × π/3 = 2π/3

- The arc AB is in the half of the circle then there are another arc

with the same length in the other half of the circle

- The length of the arc we must excluded from the length of the

circle to have the part of length of AB is greater than 2 is

2(2π/3) = 4π/3

∴ The length of the excluded arc is 4π/3

∵ The length of the circle is 2πr

∵ r = 2

∴ The length of the circle = 2π(2) = 4π

- The required arc = length circle - length excluded arc

∵ The length of the excluded arc = 4π/3

∵ The length of the circle = 4π

∴ The required arc = 4π - 4π/3 = 8π/3

- Probability = required length of arc / circumference

∴ Probability = (8π/3)/(4π) = 2/3

∴ The probability that the length of chord AB > 2 cm is 2/3

4 0

3 years ago

If you will solve this math 50 point + brainlists=5+6+6+83+15×53____84​

If you will solve this math 50 point + brainlists=5+6+6+83+15×53____84