What is the answer to this equation ? 2^3-|-4|

Reparametrize the curve with respect to arc length measured from the point where t = 0 in the direction of increasing t. (Enter

jolli1 [7]

Answer:

the answer is incomplete, below is the complete question

"Reparametrize the curve with respect to arc length measured from the point where t = 0 in the direction of increasing t. (Enter your answer in terms of s.) r(t) = 3ti + (1 - 4t)j + (1 + 2t)k r(t(s)) ="

answer

$r(t(s)) = \frac{3s}{\sqrt{29} } i + (1 -\frac{4s}{\sqrt{29} }t)j + (1 + \frac{2s}{\sqrt{29} })k$

Step-by-step explanation:

The step by step procedure is to first determine the differentiate the given vector function

r(t) = 3ti + (1 - 4t)j + (1 + 2t)k

$\frac{d(r(t) = 3ti + (1 - 4t)j + (1 + 2t)k)}{dt} \\r'(t)=3i-4j+2k\\$

since s(t) is the arc length for r(t), which is define as

$s(t)=\int\limits^t_0 {||r'(t)||} \, dt$

if we substitute the value of r'(t) we arrive at

$s(t)=\int\limits^t_0 {||r'(t)||} \, dt\\s(t)=\int\limits^t_0 {\sqrt{3^{2} +4^{2}+2^{2}} \, dt\\s(t)=\int\limits^t_ 0{\sqrt{29} } \, dx\\$

$s(t)=\int\limits^t_ 0{\sqrt{29} } \, dx\\\\s(t)=\sqrt{29} t\\hence \\t(s)=\frac{s}{\sqrt{29} }$

substituting the value of t in to the given vector equation we have

$r(t(s)) = \frac{3s}{\sqrt{29} } i + (1 -\frac{4s}{\sqrt{29} }t)j + (1 + \frac{2s}{\sqrt{29} })k$

4 0

4 years ago

Which of the following are solutions to the equation below?

kirill115 [55]

Answer: B and C

Explanation:
x^2 + 3x - 18 = 0
(x + 6)(x - 3) = 0

Set x + 6 = 0 => x = -6

Set x - 3 = 0 => x = 3

6 0

3 years ago

a quadratic equation has zero real number solutions. which could be the discriminant value associated with this equation? a. –5

Flauer [41]

Answer: Option 'a' is correct.

Step-by-step explanation:

As we know about the discriminant value i.e.

if we have given a quadratic equation :

$ax^2+bx+c=0$

then,

discriminant value will be

$b^2-4ac$

There are 3 properties regarding discriminant value as follows:

$1)b^2-4ac>0\\\\means,\\\\\text{ there exists real zeros}\\\\2)b^2-4ac=0\\\\means,\\\\\text{ there exists equal zeroes}\\\\3)b^2-4ac$

So, here we need a zero real number solutions i.e. there does not exists any real number.

So,

$b^2-4ac=-5$

Hence, Option'a' is correct.

5 0

4 years ago

Read 2 more answers

What is y=2x+4 located on the coordinate plane?

Andrei [34K]

Can I have brainliest please

5 0

2 years ago

Read 2 more answers

Answer the precalculus question from khan academy pls! I will give brainliest and 5 stars. 30 points.

rosijanka [135]

The value of f(h(2)) =2 and h(f(16))= 14

<h3>What is function?</h3>

Functions are the fundamental part of the calculus in mathematics. The functions are the special types of relations. A function in math is visualized as a rule, which gives a unique output for every input . Mapping or transformation is used to denote a function in math. These functions are usually denoted by letters . The domain is defined as the set of all the values that the function can input while it can be defined. The range is all the values that come out as the output of the function involved.

given:

f(x)= √x-1 , h(x)= x² + 5

Now,

f(h(2))= f( (2)² +5 )

=f(4+5)

=f(9)

=√9-1

= 3-1

=2

h(f(16)) = h( √16-1)

=h( 4-1)

=h(3)

=3² + 5

=9+5

=14

Learn more about function here:

brainly.com/question/12431044

#SPJ1

7 0

2 years ago