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vaieri [72.5K]
3 years ago
14

At her birthday​ party, Ms. Willow would not give her age directly. She​ said, 'If you add the year of my birth to this​ year, s

ubtract the year of my 25th birthday and the year of my 55th ​birthday, and then add my present​ age, the result is 58​.
Mathematics
1 answer:
Nimfa-mama [501]3 years ago
5 0

Answer:

69 years

Step-by-step explanation:

Given that:

Let year of birth = x

This year = 2020

Year of 25th birthday = x + 25

Year of 55th birthday = x + 55

Present age = 2020 - x

Hence,

x + 2020 - (x + 25 + x + 55) + (2020 - x) = 58

x + 2020 - (2x + 80) + (2020 - x) = 58

x + 2020 - 2x - 80 + 2020 - x = 58

-2x + 4040 - 80 = 58

-2x + 3960 = 58

-2x = 58 - 3960

-2x = - 3902

x = 1951

Birth year = 1951

,

Hence, age =

2020 - 1951 = 69 years

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What is the sum or difference 2/3(7w+4) - 1/3(2w-1)
IRISSAK [1]

Answer:

4w+3

Step-by-step explanation:

2/3(7w+4)-1/3(2w-1)

14/3w+8/3-2/3w+1/3

14/3w-2/3w+8/3+1/3

12/3w+9/3

4w+3

8 0
3 years ago
Can you write 2x(x-1) as a product and as a sum please? Thank you
kotykmax [81]

Answer: yes you can actually.

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Its a math test again-
Step2247 [10]

Answer:

A. t > 2.5

Step-by-step explanation:

If she wants to achieve <em>more </em>than 590, we already know the sign would be greater than (>).

All we have to do is divide 590/236, which gives us 2.5.

Therefore, she will have to ride more than 2.5 hours to reach her goal, t > 2.5

Hope this helps :)

Good luck on your test.

3 0
3 years ago
What are the first, second, and third frequencies audible from a 20 cm long organ pipe when (A) only one end is open and when (B
kotykmax [81]
A) The answers are:
the first frequency - 428.75 Hz
the second frequency - 1286.25 Hz
the third frequency - 2143.75 Hz

The frequency (when the pipe is closed) is: f = v(2n - 1)/4L
v - the speed of sound
n - the frequency order
L - the length of the organ pipe

We know:
v = 343 m/s
L = 20 cm = 0.2 m

1. The first frequency (n = 1):
f = 343 * (2 * 1 - 1) / 4 * 0.2 = 343 * 1 / 0.8 = 428.75 Hz


2. The second frequency (n = 2):
f = 343 * (2 * 2 - 1) / 4 * 0.2 = 343 * 3 / 0.8 = 1286.25 Hz


3. The third frequency (n = 3):
f = 343 * (2 * 3 - 1) / 4 * 0.2 = 343 * 5 / 0.8 = 2143.75 Hz


B) The answers are:

the first frequency - 857.5 Hz
the second frequency - 1715 Hz
the third frequency - 2572.5 Hz

The frequency (when the pipe is open) is: f = vn/2L
v - the speed of sound
n - the frequency order
L - the length of the organ pipe

We know:
v = 343 m/s
L = 20 cm = 0.2 m

1. The first frequency (n = 1):
f = 343 * 1 / 2 * 0.2 = 343 / 0.4 = 857.5 Hz


2. The second frequency (n = 2):
f = 343 * 2 / 2 * 0.2 = 686 / 0.4 = 1715 Hz


3. The third frequency (n = 3):
f = 343 * 3 / 2 * 0.2 = 1029 / 0.4 = 2572.5 Hz
4 0
3 years ago
296.32 divided by 4= how to solve<br> pls
astraxan [27]

Hello I hope you are having a Great day :)

Your question, 296.32 divided by 4

Your answer would be 74.08

Hopefully that helps you :)

7 0
3 years ago
Read 2 more answers
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