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3 years ago

PLZ I NEED ANSWERS NOW!!!! {(-1, 2), (0, 2), (5, 2)} is a function. TrueFalse

Mathematics

1 answer:

KIM [24]3 years ago

7 0

Answer:

True

Step-by-step explanation:

every x only has one y

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PERSON WHO GETS THIS RIGHT WILL GET A LOTTT OF POINTS 60+

jarptica [38.1K]

Answer:

See below.

Step-by-step explanation:

x = cost of adult ticket

y = cost of student ticket

cost of adult ticket is twice the cost of a student ticket

x = 2y

number of adult tickets = 64

number of student tickets = 132

cost of all adult tickets = 64x

cost of all student tickets = 132y

cost of all adult and student tickets combined: 64x + 132y

cost of all adult and student tickets combined: 1040

This gives us the equation:

64x + 132y = 1040

Use substitution to solve the system of equations. Since x = 2y, where you see x in the second equation, substitute it with 2y.

64(2y) + 132y = 1040

128y + 132y = 1040

260y = 1040

y = 1040/260

y = 4

x = 2y = 2(4) = 8

adult ticket: $8

student ticket: $4

5 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

How do you do this?

zaharov [31]

4sin x=2sin x + √3
4sin x-2sinx=√3
2sin x=√3
sinx=√3/2

x=arcsin √3/2=π/3 + 2Kπ U 2π/3+2Kπ

Sol: π/3 + 2Kπ U 2π/3+2Kπ ; K∈Z

π/3+2Kπ=60º+360ºk
2π/3+2Kπ=120º+360ºK

4 0

3 years ago

A student used this complex fraction to find a unit rate: (340 pages)/(3/4 hour.) Which quotient equals the fraction?

KatRina [158]

Answer:

Step-by-step explanation:

3 0

3 years ago

In the accompanying diagram of circle O, chords AB and CD intersect at E. If AE = 3, EB = 4, CE = x, and ED = x + 1, find. CE.

NemiM [27]

Answer:

CE = 3

Step-by-step explanation:

Given:

AE = 3

EB = 4

CE = x

ED = x + 1

Required:

Solution:

First we need to determine the value of x by generating am equation as follows:

CE × ED = AE × EB (Interacting Chords Theorem)

Substitute

x × (x + 1) = 3 × 4

x² + x = 12

Subtract 12 from each side

x² + x - 12 = 12 - 12

x² + x - 12 = 0

Factorize to find the value of x

x² + 4x - 3x - 12 = 0

x(x + 4) -3(x + 4) = 0

(x - 3)(x + 4) = 0

x - 3 = 0

x = 3

x + 4 = 0

x = -4

Let's use the positive value of x

CE = x = 3

8 0

3 years ago