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sveticcg [70]
3 years ago
10

PLEASE PLEASE PLEASE HELP ME​

Mathematics
1 answer:
qaws [65]3 years ago
5 0

Answer:

\large\boxed{\dfrac{1}{45}}

Step-by-step explanation:

1.

2 yellow marables of 10 all marables

P(A_1)=\dfrac{2}{10}=\dfrac{1}{5}

2.

1 yellow marable of 9 all marables

P(A_2)=\dfrac{1}{9}

P(A)=P(A_1)\cdot P(A_2)\\\\P(A)=\dfrac{1}{5}\cdot\dfrac{1}{9}=\dfrac{1}{45}

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a box 12 cm long, 5 cm wide and 12 cm height. A cardboard rectangle is inserted along the diagonal to divide the box vertically
Ipatiy [6.2K]

Answer:

The Cardboard dimensions are 13cm in length and 12cm in breadth

Step-by-step explanation:

We can use Pythagoras Theorem to find the length the Cardboard (a²+b²=c²). Let A be 5 and B be 12, we find that C is equivalent to 13. The height is 12, thus the breadth of the Cardboard is also 12. Hope this helps :)

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3 years ago
How can I solve this inequality?
Romashka-Z-Leto [24]
Subtract 3 from the right and do the same to the left which leaves you to x = -1
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5 0
3 years ago
Someone help me with this
puteri [66]

Answer:

A) t = 1 second to reach the highest point

B) h = 96 feet

C) t = 3.449 seconds to hit the ground

Step-by-step explanation:

We notice that the equation that describes the position of the object in terms of the time, is a parabola (quadratic) with negative leading coefficient. therefore this is a parabola with arms pointing down, and with vertex at the top. It is at that top vertex that the maximum altitude of the object is reached.

We need therefore to find the position of the vertex (time "t" is the horizontal coordinate and height "h" is the vertical coordinate).

We know that the vertex (maximum of a parabola of this type - normally described by y=ax^2+bx+c) is given by x_{vertex} = -\frac{b}{2*a}. Therefore in our case, understanding that time "t" is equivalent to the variable "x", and eight "h" is equivalent to the variable "y", that "-16" is a, and "32" is b, we have that the maximum occurs for:

t_{max}=-\frac{32}{2*(-16)} = \frac{-32}{-32} = 1

which means at one second from the launching.

We then can find the answer to part B by replacing t with "1" second in the formula for the height "h":

h(1)=-16*(1)^2+32(1)+80=-16+32+80=96

That is : a height of 96 feet.

To find the answer for part C: the time to hit the ground, we solve for the variable "t" in the given expression for the height, by setting the height to zero (object touching the ground) and using the quadratic formula:

0=-16t^2+32t+80\\t=\frac{-32+/-\sqrt{32^2-4*(-16)*80} }{2*(-16)} \\ t=\frac{-32+/-\sqrt{6144} }{-32}

which gives us two solutions: t = -1.449 seconds, and t = 3.449 seconds

Since the negative time does not have physical meaning in our case (time before the object was launched), we adopt the second answer: t = 3.449 seconds

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The LCM of 24 and 72 is 72 :)

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