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2 years ago

The value of x must be greater than _____

Mathematics

1 answer:

bezimeni [28]2 years ago

8 0

Answer: The value of x must be greater than 3

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Explanation:

The triangle inequality theorem will come into play here. The basic idea of this theorem is "take any two sides of a triangle and add them up. That sum must be larger than the third side".

So if we have
a = 12
b = x
c = 15

then the following must hold true (all three inequalities must be true)
a+b > c
a+c > b
b+c > a

Focus on the first inequality and plug in the given values. Then solve for x
a+b > c
12+x > 15
12+x-12 > 15-12
x > 3

So we see that x > 3. Repeat the same for the second inequality
a+c > b
12+15 > x
27 > x
x < 27

Repeat again for the third inequality
b+c > a
x+15 > 12
x+15-15 > 12-15
x > -3

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In summary so far, we have
x > 3
x < 27
x > -3

Combine all of those inequalities to form one single compound inequality which is 3 < x < 27

For some reason your teacher doesn't want you to focus on the "less than 27" part, so it seems like s/he only wants the "x > 3" portion.

So this is why x must be larger than 3 (up only til you get to 27 though).

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Answer:

Step-by-step explanation:

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sukhopar [10]

Answer:

The required equation is:

$y = -\frac{4}{3}t^2 -\frac{8}{3}t + 4$

Explanation:

Let us assume that the hole is at y = 0m, with x as the time.

From the question we have (-1s, 8m) as the vertex (here x being the time variable is supposed to be in seconds and y being the distance variable is supposed to be in meters)

At x = 1s, the ball gets to the hole, therefore we have point (1s, 0m)

We know that the vertex of the parabola y = ax² + bx + c is at

$x =\frac{-b}{2a}$

therefore we have:

$-1 = \frac{-b}{2a}$

We then have the following equations:

$8 = a\times -1^2 + b\times -1 + c$

$0 = a\times -1^2 + b\times 1 + c$

$-1 = \frac{-b}{2a}$

From the 3rd equation we have

1 X 2a = b.

Therefore we have:

$8 = a\times -1^2 - 1\times 2a\times1 + c$

$0 = a\times 1^2 + 1 \times2a\times 1 + c$

We can simplify both equations and get:

$8 = a\times( -1^2 - 2s^2) + c = -a\times 3^2 + c$

$0 = a\times(1^2 + 2^2) + c = a\times 3^2 + c$

The first equation now becomes:

$8 = -a\times 3 - a\times 3 = -a\times 6$

$a = frac{8}{-6} = -\frac{4}{3}$

With a, we can find the values of c and b.

$c = -a\times3 = -(-\frac{4}{3})*3 = 4$

$b = 1\times 2a = 1\times 2(-\frac{4}{3})= -\frac{8}{3}$

Then the equation is:

$y = -\frac{4}{3}t^2 -\frac{8}{3}t + 4$

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