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Alexxx [7]
1 year ago
15

Train A and train B stops at Swindon at 10:30 . Train A stops every twelve minutes and train B stops every 14 Mins , when do the

y both stop at Swindon again ?
Mathematics
1 answer:
Nata [24]1 year ago
4 0

Answer: at 11:54

Step-by-step explanation:

Let's define the 10:30 as our t = 0 min.

We know that Train A stops every 12 mins, and Train B stops every 14 mins, they will stop at the same time in the least common multiple of 12 and 14.

To find the least common multiple of two numbers, we must do:

LCM(a,b) = a*b/GCD(a,b)

Where GCD(a, b) is the greatest common divisor of a and b.

In this case the only common divisior of 12 and 14 is 2.

So we have:

LCM(12, 14) = 12*14/2 = 84.

Then the both trains will stop 84 minutes after 10:30

one hour has 60 mins, so we can write 84 minutes as:

1 hour and 24 minutes = 1:24

Then they will stop at the same time at 10:30 + 1:24 = 11:54

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Triangle XYZ with vertices X(0, 0), Y(0, –2), and Z(–2, –2) is rotated to create the image triangle X'(0, 0), Y'(2, 0), and Z'(2
Marina86 [1]

Answer: The correct options are

(A) Rotation 90° anticlockwise.

(D)  (x, y) → (–y, x).

Step-by-step explanation:  Given that ΔXYZ is rotated to create the image triangle ΔX'Y'Z'.

Triangle XYZ and its image triangle X'Y'Z' are shown in the attached figure.

The co-ordinates of the vertices of ΔXYZ are X(0, 0), Y(0, -2) and Z(-2, -2).

And the co-ordinates of the vertices ΔX'Y'Z' are X'(0, 0), Y'(2, 0) and Z'(2, -2).

<u>Option (A) Rotation 90°:</u>

We see from the figure that if we rotate ΔXYZ is rotated 90° anticlockwise, then it will coincide with ΔX'Y'Z'.

So, rotation of 90° anticlockwise is a correct option.

<u>Option (B) Rotation 180°:</u>

If we rotate ΔXYZ is rotated clockwise or anticlockwise 180°, then it will NOT coincide with ΔX'Y'Z'.

So, rotation of 180° is NOT a correct option.

<u>Option (C) Rotation 270°:</u>

If we rotate ΔXYZ is rotated clockwise 270°, then also it will not coincide with ΔX'Y'Z'.

So, rotation of 270° clockwise is also a correct option.

<u>Option (D) (x, y) → (–y, x):</u>

We see that the co-ordinates of both the triangle follow the transformation

X(0, 0)   ⇒  X'(0, 0)

Y(0, -2)  ⇒   Y'(2, 0)

Z(-2, -2)  ⇒   Z'(2, -2).

So, the transformation is (x, y) ⇒  (-y, x).

Therefore, the  transformation (x, y) → (–y, x) is a correct option.

<u>Option (E) (x, y) → (y, -x):</u>

We see that the co-ordinates of both the triangle does NOT follow this transformation

For example, suppose this transformation is correct. Then, we have

Y(0, -2)  ⇒  (-2, 0), which are not the co-ordinates of Y'.

Therefore, the  transformation (x, y) → (–y, x) is NOT a correct option.

Thus, the correct options are:

(A) Rotation 90° anticlockwise.

(D)  (x, y) → (–y, x).

9 0
1 year ago
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must a function that is decreasing over a given interval always be negative over that same interval explain
kolbaska11 [484]

check the picture below.


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8 0
2 years ago
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James is four years younger than Austin. If three times James' age is increased by the square of Austin's age, the result is 28.
lara31 [8.8K]
Let's use J for James's age and A for Austin's age. The equations are:

J = A - 4
3J + A² = 28

Just plug (A - 4) in the place of J in the second equation. This gives you:

3(A - 4) + A² = 28
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A² + 3A - 12 = 28
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A = 5 or -8

-8 is nonsense, so Austin is 5 years old. Therefore, James is 1 year old.
7 0
1 year ago
Can I have help on this question, please?
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Answer:

p(a)=3/4

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6 0
2 years ago
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The square of a number decreased by 3 times the number 28 find all possible values for the number
stealth61 [152]

Question:

The square of a number decreased by 3 times the number is 28 find all possible values for the number  

Answer:

The possible values of number are 7 and -4

Solution:

Given that the square of a number decreased by 3 times the number is 28

To find: all possible values of number

Let "a" be the unknown number

From given information,

square of a number decreased by 3 times the number = 28

a^2 - 3a = 28

a^2 - 3a - 28 = 0

Let us solve the above quadratic equation

\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Using the above formula,

\text { For } a^{2}-3 a-28=0 \text { we have } a=1, b=-3, c=-28

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Thus the possible values of number are 7 and -4

5 0
1 year ago
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