All categories

3 years ago

If now Lina is three times as old as Nick, and in 6 years she will be twice as old as he, how old are they now?\

Mathematics

1 answer:

Kitty [74]3 years ago

8 0

Answer:

Step-by-step explanation:

Nick's age = x years

Lina's age =3*x = 3x

After 6 years,

Lina's age = 3x + 6

Nick's age = x + 6

3x + 6 = 2*(x+6)

3x + 6 = 2*x + 2*6

3x + 6 = 2x + 12 {Subtract 6 form both sides}

3x +6 - 6 = 2x + 12 - 6

3x = 2x + 6 {subtract 2x from both sides}

3x - 2x = 2x + 6 - 2x

x = 6

Nick's age = 6 years

Lina's age =3*6 = 18 years

You might be interested in

The graphs below have the same shape. What is the equation of the blue graph?

Harlamova29_29 [7]

Answer:

for this graph the answer is D

7 0

3 years ago

2X+15= -3x how do you work this equation

Sphinxa [80]

Answer: $x=-3$

Add 3x to both sides

$2x+15+3x=-3x+3x\\5x+15=0$

Subtract 15 from both sides

$5x+15-15=0-15\\5x=-15$

Divide both sides by 5

$5x/5=-15/5\\x=-3$

8 0

3 years ago

Read 2 more answers

Marcel said he can find 52 times 10 by adding a 0 to the end of 52 to get 520

Rainbow [258]

Answer:

No he can not because if you add 0 to 52 you get 520 and the answer has to be 520

Step-by-step explanation:

4 0

3 years ago

What is the length of the arc if: 11. r=10 n=20 A15(pi)/ 7 B13(pi)/ 5 C16(pi)/ 2 D11(pi)/ 4 E 10(pi)/ 9 F 9(pi)/ 4 12. r=3 n=6 A

Vilka [71]

Step-by-step explanation:

The formula for arc length [for the angle in degrees] is:

$L = 2\pi r \left(\dfrac{n}{360}\right)$

here,

$n$ = degrees

$r$ = radius

using this we'll solve all the parts:

r = 10, n = 20:

$L = 2\pi r \left(\dfrac{n}{360}\right)$

$L = 2\pi (10) \left(\dfrac{20}{360}\right)$

from here, it is just simplification:

2 and 360 can be resolved: 360 divided by 2 = 180

$L = \pi (10) \left(\dfrac{20}{180}\right)$

10 and 180 can be resolved: 180 divided by 10 = 18

$L = \pi \left(\dfrac{20}{18}\right)$

finally, both 20 and 18 are multiples of 2 and can be resolved:

$L = \pi \left(\dfrac{10}{9}\right)$

$L = \dfrac{10\pi}{9}$ Option (E)

r=3, n=6:

$L = 2\pi r \left(\dfrac{n}{360}\right)$

$L = 2\pi (3) \left(\dfrac{6}{360}\right)$

$L = \dfrac{\pi}{10}$ Option (D)

r=4 n=7

$L = 2\pi r \left(\dfrac{n}{360}\right)$

$L = 2\pi (4) \left(\dfrac{7}{360}\right)$

$L = \dfrac{7\pi}{45}$ Option (C)

r=2 n=x

$L = 2\pi r \left(\dfrac{n}{360}\right)$

$L = 2\pi (2) \left(\dfrac{x}{360}\right)$

$L = \dfrac{x\pi}{90}$ Option (D)

r=y n=x

$L = 2\pi r \left(\dfrac{n}{360}\right)$

$L = 2\pi (y) \left(\dfrac{x}{360}\right)$

$L = \dfrac{xy\pi}{180}$ Option (E)

6 0

3 years ago

A hiker discovers a skull that contains 41% of its original amount of C-14. Find the age of the skull to the nearest year

snow_tiger [21]

We need the half-life of C-14 which is 5,730 years.
Now, we will need a half-life equation:
elapsed time = half-life * log (bgng amt / ending amt) / log 2
We'll say beginning amount = 100 and ending amount = 41
elapsed time = 5,730 * log (100/41) / log 2
elapsed time = 5,730 * log ( 2.4390243902 ) / 0.30102999566
elapsed time = 5,730 * 0.38721614327 / 0.30102999566
elapsed time = 5,730 * 1.2863041851 
elapsed time = 7,370.523 years

Source:
http://www.1728.org/halflife.htm

3 0

3 years ago