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3 years ago

The ratio of Wei Ling's age to Wei Xuan's age 5 years ago was 2: 5. In 9 years

Mathematics

1 answer:

Archy [21]3 years ago

6 0

Answer:

15 years old

Step-by-step explanation:

Start by defining the variables that we are going to use throughout our working:

Let the current age of Wei Ling and Wei Xuan be L and X years old respectively.

Next, form equations using the given information.

5 years ago

Wei Ling: (L -5) years old

Wei Xuan: (X -5) years old

Given that the ratio of Wei Ling's age to that of Wei Xuan's is 2: 5,

$\frac{ L - 5}{X - 5} = \frac{2}{5}$

Cross multiply:

2(X -5)= 5(L -5)

Expand:

2X -10= 5L -25

2X= 5L -25 +10

2X= 5L -15 -----(1)

9 years time

Wei Ling: (L +9) years old

Wei Xuan: (X +9) years old

Given that the ratio of Wei Ling's age to that of Wei Xuan is 3: 4,

$\frac{L + 9}{X + 9} = \frac{3}{4}$

Cross multiply:

3(X +9)= 4(L +9)

Expand:

3X +27= 4L +36

3X= 4L +36 -27

3X= 4L +9 -----(2)

Let's solve using the elimination method.

(1) ×3:

6X= 15L -45 -----(3)

(2) ×2:

6X= 8L +18 -----(4)

(3) -(4):

6X -6X= 15L -45 -(8L +18)

0= 15L -45 -8L -18

0= 7L -63

7L= 63

L= 63 ÷7

L= 9

Substitute L= 9 into (1):

2X= 5(9) -15

2X= 45 -15

2X= 30

X= 30 ÷2

X= 15

Thus, Wei Xuan is 15 years old now.

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3 years ago

Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

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Vladimir79 [104]

Answer:

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