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3 years ago

Which set of points are collinear? P, T, Y D, S, Y H, T, S Y, D, H

Mathematics

1 answer:

Marizza181 [45]3 years ago

7 0

Answer: P, T, Y

This is because the points on the line are straight

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Find the output values of the exponential function g(x) = 4,096x for x = 0, 0.25, 0.50, 0.75, and 1.

WARRIOR [948]

Hello,

g(x)=4096^x

g(0)=1

g(0.25)=(2^12)^1/4=2^(12/4)=2^3=8

g(0.50)=(2^12)^(1/2)=2^6=64

g(0.75)=(2^12)^(3/4)=2^9=512

g(1)=4096^1=4096

6 0

3 years ago

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What is the absolute value of -77?

eimsori [14]

$|x|=\begin{cases}x&\text{for }x\ge0\\[1ex]-x&\text{for }x$

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3 years ago

Solve the system of equations. x + 3y = −12x + 2y = 6

UkoKoshka [18]

Answer:

x=-3/19

y=39/19

Step-by-step explanation:

the equations can be written as the following system

$x+3y=6\\-12x+2y=6$

If the second equation is multiplied by 3/2 it will become -18x+3y=9

If that equation is then subtracted with the first you will obtain the following

-19x=3, it means x=-3/19

The other variable is obtained by substituting the new value for x in the first equationx+3y=6, thus we obtain that y=39/19

7 0

3 years ago

Q1 A ball is thrown upwards with some initial speed. It goes up to a height of 19.6m and then returns. Find (a) The initial spee

lubasha [3.4K]

Answer:

(a) 19.6 ms⁻¹

(b) 2 s

(d) 4 s

Step-by-step explanation:

<u>Constant Acceleration Equations (SUVAT)</u>

$\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}$

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

Acceleration due to gravity = 9.8 ms⁻².

When the ball reaches its maximum height, its velocity will momentarily be zero.

<u>Given values</u> (taking up as positive):

$s=19.6 \quad v=0 \quad a=-9.8$

$\begin{aligned}\textsf{Using} \quad v^2&=u^2+2as\\\\\textsf{Substitute the given values:}\\0^2&=u^2+2(-9.8)(19.6)\\0&=u^2-384.16\\u^2&=384.16\\u&=\sqrt{384.16}\\\implies u&=19.6\; \sf ms^{-1}\end{aligned}$

Therefore, the initial speed is 19.6 ms⁻¹.

Using the same values as for part (a):

$\begin{aligned}\textsf{Using} \quad s&=vt-\dfrac{1}{2}at^2\\\\\textsf{Substitute the given values:}\\19.6&=0(t)-\dfrac{1}{2}(-9.8)t^2\\19.6&=4.9t^2\\t^2&=\dfrac{19.6}{4.9}\\t^2&=4\\t&=\sqrt{4}\\\implies t&=2\; \sf s\end{aligned}$

Therefore, the time taken to reach the highest point is 2 seconds.

As the ball reaches its maximum height at 2 seconds, one second before this time is 1 s.

<u>Given values</u> (taking up as positive):

$u=19.6 \quad a=-9.8 \quad t=1$

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=19.6+(-9.8)(1)\\v&=19.6-9.8\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

The velocity of the ball one second before it reaches its maximum height is the <u>same</u> as the velocity one second after.

<u>Proof</u>

When the ball reaches its maximum height, its velocity is zero.

Therefore, the values for the downwards journey (from when it reaches its maximum height):

$u=0 \quad a=9.8 \quad t=1$

(acceleration is now positive as we are taking ↓ as positive).

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=0+9.8(1)\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

Therefore, the velocity of the ball one second before <u>and</u> one second after it reaches the maximum height is 9.8 ms⁻¹.

From part (a) we know that the time taken to reach the highest point is 2 seconds. Therefore, the time taken by the ball to travel from the highest point to its original position will also be 2 seconds.

Therefore, the total time taken by the ball to return to its original position after it is thrown upwards is 4 seconds.

4 0

1 year ago

Helpppppppppp.......

Andrew [12]

Answer:

36cm

Step-by-step explanation:

The scale is 3 to 1, so the poster size is 3 times larger than the actual thing.

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3 years ago

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