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3 years ago

Jamie is in charge of covering the display box in felt including the bottom what is the total area to be covered

Mathematics

1 answer:

Aliun [14]3 years ago

6 0

Answer:

1,260

Step-by-step explanation:

I know I'm right for sure

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Solve x2 – 64 = 0. Isolate x2: x2 = 64 Apply the square root property of equality: Isolate the variable: x = x =

dybincka [34]

Solve the equation by isolating the variable you're solving for on one side of the equation and get everything on the opposite side.

$x^2 - 64 = 0$

The x variable is being square and is being subtracted by 64.
Find the inverse of all of these operations.

First, get rid of the subtraction operation. The inverse of subtraction is addition.

Add 64 to both sides of the equation.

$x^2 - 64 + 64 = 0 + 64 \\ x^2 = 64$

Now the variable x is only being squared. Reverse this operation by using the inverse of exponents.

Take the square root of both sides.

$\sqrt{x^2} = \sqrt{64} \\ x = 8$

So, x is equal to 8.

6 0

4 years ago

Read 2 more answers

Find the x and y intercepts <br> 4x-5y=40

IrinaVladis [17]

Answer:

x intercept

$y = 0 \\ 4x - 5(0) = 40 \\ 4x = 40 \\ x = 10$

y intercept

$x = 0 \\ 4(0) - 5y = 40 \\ - 5y = 40 \\ y = - 8$

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4 years ago

Where could you insert parentheses in The problem 310-20÷2×5+13 so the answer should be 130

Inessa [10]

The parenthesis should be placed in 310-20/2 x (5 + 13) =130

3 0

3 years ago

3x-9=-21 equation in words

sergeinik [125]

'nine less than three times a number is negative twenty-one.'

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4 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