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nevsk [136]
2 years ago
11

2zx2y=18 where x=1,z=y

Mathematics
2 answers:
Bad White [126]2 years ago
5 0

\huge \bf༆ Answer ༄

Let's plug the given values and find value of y ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2zx2y = 18

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2 \times y \times 1 \times 2 \times y = 18

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:4 {y}^{2}  = 18

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:{y}^{2}  =  \dfrac{18}{4}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y {}^{2}  = 4.5

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y =  \sqrt{4.5}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y =  \sqrt{9 \times 0.5}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y = 3 \sqrt{ \frac{5}{10} }

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:3 \sqrt{ \frac{1}{2} }

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: \dfrac{3}{ \sqrt{2} }

or, you can further simplify it to ;

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2.12 \:  \:  (upto \:  \: two \: decimals)

QveST [7]2 years ago
3 0
Plug in the values: 2(y)(1)2y=18
First multiply 1 with 2y = 2y(2y)=18
Second multiply 2y and 2y = 4y^2=18
Third divide both sides by 4 = 4y^2/4=18/4
—-> y^2=9/2
Fourth square root both sides = y= √ 9/2 (in square root)
If you need to simplify further you would get 3 √2/2
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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

Or

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.

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12x⁴y²divided by 3x³y to the 5 th power....Please help me out....
ki77a [65]
To divide the expression we proceed as follows:
12x⁴y²÷3x³y⁵
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