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3 years ago

I’m stuck, need help, please?

Mathematics

1 answer:

Ymorist [56]3 years ago

4 0

Answer:

C. Line 3

Step-by-step explanation:

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= 3x +15 <br> = 4x-10<br><br> What is the value of x?

Viefleur [7K]

Answer:

3x+15= 4x-10

3x-4x= -10-15

-x= -25(minus will be cut on both sides)

so x= 25

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3 years ago

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vekshin1

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3 years ago

Which method do you prefer to use to find sums

VMariaS [17]

Usually you use addition to find a sum.

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3 years ago

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21. Given that a line y = x passes through the origin and up to the right, what would the line y = mx look like if m = 1/2?

maks197457 [2]

a is the answer as the only change is the gradient being halved,

b would be if it was -1/2 and c would be1/2 x+1/2

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3 years ago

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