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2 years ago

Order these numbers in least to greatest (THIS IS NOT A TEST OR ASSIGNMENT) I do need help ASAP and I WILL GIVE BRAINLIEST

Mathematics

2 answers:

Pie2 years ago

7 0

Answer:

1. LESS THAN ( 6.4... < 6.8)

2. GREATER THAN ( -6.4... > -6.8)

3. GREATER THAN (1.44 > -1.44)

4. LESS THAN (-3.1... < -3)

Step-by-step explanation:

Misha Larkins [42]2 years ago

5 0

Answer:

1. <

2. >

3. >

4.<

least to greatest:

-6.8< $-\sqrt{41}$ < $-\sqrt{10}$ < -3< $-(1.2)^{2}$ < $(-1.2)^{2}$ < $\sqrt{41}$ < 6.8

Step-by-step explanation:

1. 6.4 and 6.8

2. -6.4 and -6.8

3. 1.44 and -1.44

4. -3.16 and -3

the third one on the picture is wrong

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10/16 x 2/16 = 12/16 in multiplication form

ivann1987 [24]

Correct answer:

5/64

Step-by-step explanation:

Multiply the numerators: 10 × 2 = 20
16 × 16 = 256
Put them together: 20/256
Simplify: 20/256 = 5/64

I hope this helps!

3 0

2 years ago

(2x+5)/3=-7 show all work

Phantasy [73]

→(2X+5)=-21→2X=-26→X=-13

3 0

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:

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