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vaieri [72.5K]
3 years ago
10

Which number line correctly shows 1.5 + 2.5?

Mathematics
2 answers:
dybincka [34]3 years ago
7 0

Answer:

Should be A right?

Step-by-step explanation:

Cuz you go from 0 to 1.5 which means it is 1.5 distance and then it goes from 4 to 1.5 which means (4-1.5)= 2.5 distance?

First time seen this but makes sense to me

jek_recluse [69]3 years ago
4 0

Answer:

A

Step-by-step explanation:The answer choice A is correct because it correctly shows both numbers.All of the other answer choices didnt correctly show the answer choices,and they only showed part of the correct answers.So therfore your correct answer is A(hope this helps hmu if it doesnt)

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Solve for x. 1/15+x=3/10
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Answer:

x=7/30

Step-by-step explanation:

3/10-1/15

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Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = e−5t
Elan Coil [88]

Answer:

x = 1 - 5t

y = t

z = 1 - 5t

Step-by-step explanation:

For the equation of a line, we need a point and a direction vector. We are given a point (1, 0, 1).

Since the line is suppose to be a tangent to the given curve at the point (1, 0, 1), we need to find a tangent vector for which the curve passes through that point.

We have x = e^(-5t)cos5t

at t = 1, x = e^(-5)cos5

at t = 0, x = 1

y = e^(-5t)sin5t

at t = 1, y = e^(-5)sin5

at t = 0, y = 0

z = e^(-5t)

at t = 1, z = e^(-5)

at t = 0, z = 1

Clearly, the only parameter value for which the curve passes through the point (1, 0, 1) is t = 0.

In vector notation, the curve

r(t) = xi + yj + zk

= e^(-5t)cos5t i + e^(-5t)sin5t j + e^(-5t) k

r'(t) = [-5e^(-5t)cos5t - e^(-5t)sin5t] i +[e^(-5t)cos5t - 5e^(-5t)sin5t] j - 5e^(-5t) k

r'(0) = -5i + j - 5k

is a vector tangent at the point.

We get the parametric equation from this.

x = x(0) + tx'(0)

= 1 - 5t

y = y(0) + ty'(0)

= t

z = z(0) + tz'(0)

= 1 - 5t

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Answer:

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A. x+y<100
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