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3 years ago

Ow can 38 over 6 be expressed as a decimal?

2 answers:

5 0

38/6 as a decimal would be 6.3333 (the three is repeating)
Hope I helped.
Please mark brainilest when u get the chance :)

serious [3.7K]3 years ago

4 0

Answer:

yes, the decimal form of 38/6 would be 6.333333333 or 6.33 repeating

Step-by-step explanation:

hope that helps

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Please help me with 18, 21, 24, 27, 30, 33, 36, 39, 46, and 47!

ale4655 [162]

Answer:

18) it is parallel to y axis

7 0

3 years ago

URGENT! find the sum. the answer is not D

otez555 [7]

<h3>Answer: C) -1/2</h3>

=============================================

Explanation:

We'll apply the derivative to get

g(x) = 4x^3 + 3x^2

g ' (x) = 3*4x^2 + 2*3x

g ' (x) = 12x^2 + 6x

Set that equal to 0 and solve for x. We do this to find where the horizontal tangents are located.

g ' (x) = 0

12x^2 + 6x = 0

6x(2x + 1) = 0

6x = 0 or 2x+1 = 0

x = 0 or x = -1/2

Adding those two said solutions yields us the final answer choice C

7 0

2 years ago

Find the angle between u =the square root of 5i-8j and v =the square root of 5i+j.

fenix001 [56]

Answer:

The angle between vector $\vec{u} = 5\, \vec{i} - 8\, \vec{j}$ and $\vec{v} = 5\, \vec{i} + \, \vec{j}$ is approximately $1.21$ radians, which is equivalent to approximately $69.3^\circ$ .

Step-by-step explanation:

The angle between two vectors can be found from the ratio between:

their dot products, and
the product of their lengths.

To be precise, if $\theta$ denotes the angle between $\vec{u}$ and $\vec{v}$ (assume that $0^\circ \le \theta < 180^\circ$ or equivalently $0 \le \theta < \pi$ ,) then:

$\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}$ .

<h3>Dot product of the two vectors</h3>

The first component of $\vec{u}$ is $5$ and the first component of $\vec{v}$ is also

The second component of $\vec{u}$ is $(-8)$ while the second component of $\vec{v}$ is $1$ . The product of these two second components is $(-8) \times 1= (-8)$ .

The dot product of $\vec{u}$ and $\vec{v}$ will thus be:

$\begin{aligned} \vec{u} \cdot \vec{v} = 5 \times 5 + (-8) \times1 = 17 \end{aligned}$ .

<h3>Lengths of the two vectors</h3>

Apply the Pythagorean Theorem to both $\vec{u}$ and $\vec{v}$ :

$\| u \| = \sqrt{5^2 + (-8)^2} = \sqrt{89}$ .
$\| v \| = \sqrt{5^2 + 1^2} = \sqrt{26}$ .

<h3>Angle between the two vectors</h3>

Let $\theta$ represent the angle between $\vec{u}$ and $\vec{v}$ . Apply the formula $\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}$ to find the cosine of this angle:

$\begin{aligned} \cos(\theta)&= \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|} = \frac{17}{\sqrt{89}\cdot \sqrt{26}}\end{aligned}$ .

Since $\theta$ is the angle between two vectors, its value should be between $0\; \rm radians$ and $\pi \; \rm radians$ ( $0^\circ$ and $180^\circ$ .) That is: $0 \le \theta < \pi$ and $0^\circ \le \theta < 180^\circ$ . Apply the arccosine function (the inverse of the cosine function) to find the value of $\theta$ :

$\displaystyle \cos^{-1}\left(\frac{17}{\sqrt{89}\cdot \sqrt{26}}\right) \approx 1.21 \;\rm radians \approx 69.3^\circ$ .

3 0

3 years ago

Consider the rational expression (IMAGE ATTACHED)

Keith_Richards [23]

Answer:

3x² is a term in the numerator
x + 1 is a common factor
The denominator has 3 terms

Step-by-step explanation:

You can identify terms and count them before you start factoring. Doing so will identify 3x² as a term in the numerator, and will show you there are 3 terms in the denominator.

When you factor the expression, you get ...

$\dfrac{3x^2-3}{3x^2+2x-1}=\dfrac{3(x^2-1)}{(3x-1)(x+1)}=\dfrac{3(x-1)(x+1)}{(3x-1)(x+1)}$

This reveals a common factor of x+1.

So, the above three observations are true of this rational expression.

3 0

3 years ago

What is x - 3y = 3? any one know?

Sunny_sXe [5.5K]

Answer:

If we are solving for x it is:

x = 3y + 3

If we are solving for y it is:

x - 3 = 3y

3y = x - 3

y = (x-3)/3

3 0

3 years ago