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son4ous [18]
2 years ago
12

Please help me with this question I’m so confused

Mathematics
2 answers:
Arisa [49]2 years ago
6 0

Answer:

4.75

Step-by-step explanation:

i hope this is helpful to you.

zavuch27 [327]2 years ago
4 0

s = ut +  \frac{1}{2} a {t}^{2}  \\ s = 10( \frac{1}{2} ) +  \frac{1}{2} ( - 2) {( \frac{1}{2}) }^{2}  \\ s = 10( \frac{1}{2} ) +  \frac{1}{2} ( - 2)(  \frac{1}{4} ) \\ s = 5 + ( - 1)( \frac{1}{4} ) \\ s = 5 + ( -  \frac{1}{4} ) \\ s = 5 -  \frac{1}{4}  \\ s =  \frac{20}{4}  -  \frac{1}{4}  \\ s =  \frac{19}{4}

Hope it helps

Please give brainliest

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Answer:

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4 years ago
The water from a fire hose follows a path described by y equals 2.0 plus 0.9 x minus 0.10 x squared ​(units are in​ meters). If
Alecsey [184]

Answer:

The resultant velocity is 12.21 m/s.

Step-by-step explanation:

We are given that the water from a fire hose follows a path described by y equals 2.0 plus 0.9 x minus 0.10 x squared ​(units are in​ meters).

Also, v Subscript x is constant at 10.0 ​m/s.

The water from a fire hose follows a path described by the following equation below;

y=2.0 + 0.9x-0.10x^{2}

The velocity of the x component is constant at =  v_x=10.0 \text{ m/s}

and the point at which resultant velocity has to be calculated is (9.0,2.0).

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v_x=\frac{dx}{dt} \text{   and    }   v_y=\frac{dy}{dt}

Now, differentiating the above equation with respect to t, we get;

y=2.0 + 0.9x-0.10x^{2}

\frac{dy}{dt} =0 + 0.9\frac{dx}{dt} -(0.10\times 2)\frac{dx}{dt}

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Now, putting v_x=10.0 \text{ m/s} in the above equation;

v_y = 0.7 \times 10.0 = 7 m/s

Now, the resultant velocity is given by = v=\sqrt{v_x^{2}+v_y^{2}  }

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3 years ago
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Lana71 [14]

Answer:

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Step-by-step explanation:

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2^3 : 5^3

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