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3 years ago

Which inequality is represented by the graph ??pls help:)?

Mathematics

1 answer:

Sedbober [7]3 years ago

4 0

It’s the one to the right I hope that helps

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Paha777 [63]

Answer:

Je ne parle pas français donc j'utilise un traducteur. J'espère qu'il n'y a pas trop d'erreurs. Le frère de Maurice a dépensé 5,20 $ et Maurice 20,80 $

Step-by-step explanation:

7 0

3 years ago

a boy is standing on a pole of height 14.7m throws a stone upwards. it moves in a vertical line slightly away from the pole and

fomenos

Answer:

The time taken for the upward motion is 1 second. The same time is taken for the downward motion

It reaches a maximum height of 4.9 meters.

Step-by-step explanation:

The equation of motion is:

$x(t) = -4.9t^{2} + 9.8t$

Since the term which multiplies t squared is negative, the graph is concave down, that is, x increases until the vertex, where it reaches it's maximum height, then it decreases.

Vertex of a quadratic equation:

Quadratic equation in the format $x(t) = at^{2} + bt + c$

The vertex is the point $(t_{v}, x(t_{v}))$ , in which

$t_{v} = -\frac{b}{2a}$

In this question:

$x(t) = -4.9t^{2} + 9.8t$

So $a = -4.9, b = 9.8$

Vertex:

$t_{v} = -\frac{9.8}{2*(-4.9)} = 1$

The time taken for the upward motion is 1 second.

$x(t_{v}) = x(1) = 9.8*1 - 4.9*(1)^{2} = 4.9$

It reaches a maximum height of 4.9 meters.

Downward motion:

From the vertex to the ground.

The ground is t when x = 0. So

$-4.9t^{2} + 9.8t = 0$

$4.9t^{2} - 9.8t = 0$

$4.9t(t - 2) = 0$

$4.9t = 0$

$t = 0$

$t - 2 = 0$

$t = 2$

It reaches the ground when t = 2 seconds.

The downward motion started at the vertex, when t = 1.

So the duration of the downward motion is 2 - 1 = 1 second.

5 0

3 years ago

Whats the answer for this

Oxana [17]

Answer:

a) Angle X = 108°

<h3>C). Vertically opposite angles</h3>

as AD is parallel to EH both the angles are equal

7 0

2 years ago

Read 2 more answers

Solve this for Brainliest!<br><br>See the attachment [Level - GCSE]

madreJ [45]

Answer:

$x = 3 + 4\sqrt{3}$

Step-by-step explanation:

OPQ is a right angle triangle.

<u>Using Pythagoras</u>

$x^{2} + (x+5)^{2} = (x+8)^{2}\\x^{2} + x^{2} + 10x +25 = x^{2} + 16x + 64\\x^{2} + x^{2} - x^{2} + 10x - 16x + 25 - 64 = 0\\x^{2} - 6x -39 = 0$

<u>Using quadratic formula</u>

<u /> $x = \frac{-b \± \sqrt{b^{2}-4ac}}{2a}\\x = \frac{-(-6) \± \sqrt{(-6)^{2}-4(1)(-39)}}{2(1)}\\x = \frac{6}{2} \± \frac{\sqrt{192}}{2}\\x = 3 \± \frac{8\sqrt{3}}{2}\\x = 3 \± 4\sqrt{3}$