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3 years ago

Some students are making

Mathematics

1 answer:

Anika [276]3 years ago

8 0

Answer: 12 12/18 cups flour and 18 eggs.

Step-by-step explanation:

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find the slope and the y-intercept of the graph of the linear equation.y=−3−14xy=−3−14xthe slope isand the y-intercept is.

Vladimir79 [104]

Answer:

add the graph

Step-by-step explanation:

6 0

3 years ago

What does 3.0000÷16 equal

Bas_tet [7]

Answer:

Step-by-step explanation:

=3/16

Using Calculator

=0.1876

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4 years ago

Read 2 more answers

Complete a table with at least 5 points then graph the following quadratic on the coordinate plane. Identify all zeros.y = -2x2

Anna [14]

Answer:

See attached

Step-by-step explanation:

Given function:

y = -2x² + 5x + 8

Table and graph are attached

Zeros are included in the graph

<u>Zero's are obtained:</u>

x = 0 ⇒ y = 8

y = 0 ⇒ Solving quadratic equation

-2x² + 5x + 8 = 0
x = (-5 ± √(25 + 2*4*8))/-4
x = 3.608
x = -1.108

So zeros are (0, 8), (3.608, 0) and (-1.108, 0)

3 0

4 years ago

What is the answer for AC =

Triss [41]

Answer:

The length of AC is 2.33 units.

Step-by-step explanation:

You have to use Tangent Rule, tanθ = opposite/adjacent :

$\tan(θ) = \frac{oppo.}{adj.}$

Let oppo. = AC,

Let adj. = BC = 5,

Let θ = 25°,

$\tan(25) = \frac{ac}{5}$

$ac = 5 \tan(25)$

$ac = 2.33 \: units \: (near.hundredth)$

4 0

3 years ago

In the drawing above two rugby players start 41 meters apart at rest. Player 1 accelerates at 2.2 m/s^2 and player 2 accelerates

Anettt [7]

Answer:

The time that elapses before the players collide is 4.59 secs

Step-by-step explanation:

The distance between the two players is 41 meters. Hence, they would have total distance of 41 meters by the time they collide.

We can write that

S₁ + S₂ = 41 m

Where S₁ is the distance covered by Player 1 before collision

and S₂ is the distance covered by Player 2 before collision

From one of the equations of kinematics for linear motion,

$S = ut$ + $\frac{1}{2} at^{2}$

Where $S$ is distance

$u$ is the initial velocity

$a$ is acceleration

and $t$ is time

Since the players collide at the same time, then time spent by player 1 before collision equals time spent by player 2 before collision

That is, t₁ = t₂ = t

Where t₁ is the time spent by player 1

and t₂ is the time spent by player 2

For player 1

$S$ = S₁

$u$ = 0 m/s ( The player starts at rest)

$a$ = 2.2 m/s²

Then,

S₁ = $0(t)$ + $\frac{1}{2} (2.2)t^{2}$

S₁ = $1.1t^{2}$

$t^{2} = \frac{S_{1} }{1.1}$

For player 2

$S$ = S₂

$u$ = 0 m/s ( The player starts at rest)

$a$ = 1.7 m/s²

Then,

S₂ = $0(t)$ + $\frac{1}{2} (1.7)t^{2}$

S₂ = $0.85t^{2}$

$t^{2} = \frac{S_{2} }{0.85}$

Since the time spent by both players is equal, We can write that

$t^{2}$ = $t^{2}$

Then,

$\frac{S_{1} }{1.1} = \frac{S_{2} }{0.85}$

$0.85S_{1} = 1.1S_{2}$

From, S₁ + S₂ = 41 m

S₂ = 41 - S₁

Then,

$0.85S_{1} = 1.1 ( 41 -S_{1} )$

$0.85S_{1} = 45.1 - 1.1S_{1}$

$0.85S_{1} + 1.1S_{1} = 45.1 \\1.95S_{1} = 45.1\\S_{1} = \frac{45.1}{1.95} \\S_{1} = 23.13 m$

This is the distance covered by the first player. We can then put this value into $t^{2} = \frac{S_{1} }{1.1}$ to determine how much time elapses before the players collide.

$t^{2} = \frac{S_{1} }{1.1}$

$t^{2} = \frac{23.13 }{1.1}$

$t^{2} = 21.03\\t = \sqrt{21.03} \\t = 4.59 secs$

Hence, the time that elapses before the players collide is 4.59 secs

3 0

4 years ago