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3 years ago

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Is the expression x3•x3•x3 equivalent to x3•3•3? Why or why not? Explain your reasoning.

1 answer:

devlian [24]3 years ago

3 0

no they are different

if Base are equal power are added

like wise in first expressions power are added

but in second expressions cube of three is written as x to the power.

so they are not equal

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The length of a rectangle is 3 times the width.If the length is decreased by 4 m and the width is increased by 1 m , the perimet

o-na [289]

Answer:

Length = 27 metres

Width = 9 metres

Step-by-step explanation:

Consider the width to be x.

Then the length will be 3x.

The perimeter is 66m when the width is increased by 1m and the length is decreased by 4m.

So the dimensions will become:

Length=3x-4

Width=x+1

Now solve for the variable x using the formula for perimeter of a rectangle:

P = 2(l+b)

66 = 2(3x-4+x+1) (Divide both sides by 2 and simplify the parentheses)

33 = 4x-3 (Add 3 to both sides)

36 = 4x (Divide both sides by 4)

x = 9

Now plug-in the value of x into the original dimensions which were:

Length=3x

Width=x

So the dimensions are:

Length = 27 metres

Width = 9 metres

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2 years ago

Ethan is responsible to write the minutes for the meeting. During the meeting, he finished writing of the minutes. Before gettin

dolphi86 [110]

11/24
1. 3/8 = 9/24
2. 1/6 = 4/24
3. 9/24 + 4/24 = 13/24
4. 1- 13/24 = 11/24

And that’s if! Hoped this helped :)
These are all the correct steps to get the answer to this problem

7 0

2 years ago

Fofino [41]

Answer:

Step-by-step explanation:

A

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3 years ago

Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and

zloy xaker [14]

Answer:

$(0,0)$ $(4000,0)$ and $(500,79)$

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

$\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW$

$\frac{dW}{dt} = -0.02W + 0.00004RW$

To solve this, we first equate $\frac{dR}{dt}$ and $\frac{dW}{dt}$ to 0.

So, we have:

$0.09R(1 - 0.00025R) - 0.001RW = 0$

$-0.02W + 0.00004RW = 0$

Factor out R in $0.09R(1 - 0.00025R) - 0.001RW = 0$

$R(0.09(1 - 0.00025R) - 0.001W) = 0$

Split

$R = 0$ or $0.09(1 - 0.00025R) - 0.001W = 0$

$R = 0$ or $0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

Factor out W in $-0.02W + 0.00004RW = 0$

$W(-0.02 + 0.00004R) = 0$

Split

$W = 0$ or $-0.02 + 0.00004R = 0$

Solve for R

$-0.02 + 0.00004R = 0$

$0.00004R = 0.02$

Make R the subject

$R = \frac{0.02}{0.00004}$

$R = 500$

When $R = 500$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0$

$0.09 -0.01125 - 0.001W = 0$

$0.07875 - 0.001W = 0$

Collect like terms

$- 0.001W = -0.07875$

Solve for W

$W = \frac{-0.07875}{ - 0.001}$

$W = 78.75$

$W \approx 79$

$(R,W) \to (500,79)$

When $W = 0$ , we have:

$0.09 - 2.25 * 10^{-5}R - 0.001W = 0$

$0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0$

$0.09 - 2.25 * 10^{-5}R = 0$

Collect like terms

$- 2.25 * 10^{-5}R = -0.09$

Solve for R

$R = \frac{-0.09}{- 2.25 * 10^{-5}}$

$R = 4000$

So, we have:

$(R,W) \to (4000,0)$

When $R =0$ , we have:

$-0.02W + 0.00004RW = 0$

$-0.02W + 0.00004W*0 = 0$

$-0.02W + 0 = 0$

$-0.02W = 0$

$W=0$

So, we have:

$(R,W) \to (0,0)$

Hence, the points of equilibrium are:

$(0,0)$ $(4000,0)$ and $(500,79)$

4 0

3 years ago

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Xelga [282]

A，the intercept is equal to 70

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2 years ago

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