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2 years ago

Evaluate the expressions. -(1/3)^0 and (-9)^0

Mathematics

1 answer:

mixas84 [53]2 years ago

7 0

Answer: 1

Step-by-step explanation:

To Evaluate the expression -(1/3)^0 and (-9)^0

It is important to note that any number raised to the power of 0 is 1, therefore,

-(1/3)^0 = -1

(-9)^0 = -1

(-1) x (-1) = +1

I hope this helps

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Fill in the table and then round to the given place 225.286

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3 years ago

Solve the equation -3 = -13 - 5x for x

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Answer:

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Step-by-step explanation:

-3 = -13 - 5x

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2 years ago

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2 years ago

Two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 + 2t, 1 + 6t, 1 + 14t . Find the points at which their p

GuDViN [60]

Answer:

A) points at which paths intersect : (1,1,1) ; (2,4,8)

B) DNE

Step-by-step explanation:

A) To find the points in which the particle paths intersect, it is necessary to find the values of t for which the three components of both vectors are equal:

$t_1=1+2t_2\\\\t_1^2=1+6t_2\\\\t_1^3=1+14t_2$

you replace t1 from the first equation in the second equation:

$(1+2t_2)^2=1+6t_2\\\\1+4t_2+4t_2^2=1+6t_2\\\\4t_2^2-2t_2=0\\\\t_2(2t_2-1)=0\\\\t_2=0\\\\t_2=\frac{1}{2}$

Then, for t2 = 0 and t2=1/2 you obtain for t1:

$t_1=1+2(0)=1\\\\t_1=1+2(\frac{1}{2})=2$

Hence, for t1=1 and t2=0 the paths intersect. Furthermore, for t1=2 and t2=1/2 the paths also intersect.

The points at which the paths intersect are:

$r_1(1)=(1,1,1)=r_2(0)=(1,1,1)\\\\r_1(2)=(2,4,8)=r_2(\frac{1}{2})=(2,4,8)$

B) You have the following two trajectories of two independent particles:

$r_1(t)=(t,t^2,t^3)\\\\r_2(t)=(1+2t,1+6t,1+14t)$

To find the time in which the particles collide, it is necessary that both particles are in the same position on the same time. That is, each component of the vectors must coincide:

$t=1+2t\\\\t^2=1+6t\\\\t^3=1+14t$

From the first equation you have:

$t=1+2t\\\\t=-1$

This values does not have a physical meaning, then, the particle do not collide

answer: DNE

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2 years ago

Create a question like

vovikov84 [41]

Answer:

x = (+ - ) 9

Step-by-step explanation:

x^2 = 81

u know how u can sometimes cancel something out by doing the opposite.

This is one of those times as well. the opposite of ^2 = square root.

so if we take the square root of both sides, this will cancel out the^2 on the left side.

sqrt (x^2) = sqrt(81)

x = (+-) 9.....thats a positive and a negative sign...because the sqrt of 81 is both 9 and -9.

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2 years ago

Evaluate the expressions. -(1/3)^0 and (-9)^0​

Evaluate the expressions. -(1/3)^0 and (-9)^0