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Kamila [148]
2 years ago
15

Someone please help????

Mathematics
1 answer:
Nonamiya [84]2 years ago
7 0

Answer:

p is 13

Step-by-step explanation:

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Select the correct answer.<br> Which statement describes the situation shown in the graph?
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Xray vision, where is the graph you want answers like this?

Step-by-step explanation:

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The equation x2 – 1x – 90 = 0 has solutions {a, b}. What is a + b?
ololo11 [35]
It can be factorised as
(x - 10)(x + 9)
so the solutions are
x= 10 and x = -9

so the sum of the solutions is 10 -9 which is 1
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B
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2 years ago
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A cannonball of mass 1kg is shot vertically upward from the top of a building with an unknown velocity v_0(m/sec).v 0 ​ (m/sec).
Bumek [7]

Taking the upward direction to be positive, the cannonball's height y(t) in the air at time t is given by

y(t)=y_0+v_0 t-\dfrac g2t^2

where g is the magnitude of the acceleration due to gravity, 10 m/s^2, and y_0 is the height of the building from which the ball is being thrown.

At the moment the cannonball reaches its maximum height of 30 m, its velocity at that time is 0, so that

0^2-{v_0}^2=-2g(30\,\mathrm m-y_0)\implies v_0=\sqrt{\left(20\dfrac{\rm m}{\mathrm s^2}\right)(30\,\mathrm m-y_0)}

Substitute this into the height equation above, and let t=2\,\mathrm s, for which we have y(2\,\mathrm s)=30\,\mathrm m:

30\,\mathrm m=y_0+\sqrt{\left(20\dfrac{\rm m}{\mathrm s^2}\right)(30\,\mathrm m-y_0)}(2\,\mathrm s)-\left(5\dfrac{\rm m}{\mathrm s^2}\right)(2\,\mathrm s)^2

Solve for y_0: (units omitted for brevity; we know that y_0 should be given in m)

30=y_0+4\sqrt{150-5y_0}-20

50-y_0=4\sqrt{150-5y_0}

(50-y_0)^2=\left(4\sqrt{150-5y_0}\right)^2

2500-100y_0+{y_0}^2=16(150-5y_0)

{y_0}^2-20y_0+100=0

(y_0-10)^2=0

\implies\boxed{y_0=10\,\mathrm m}

3 0
3 years ago
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