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3 years ago

13

How do I rotate a point by 180 degrees clockwise?

1 answer:

gizmo_the_mogwai [7]3 years ago

8 0

180 degrees is half of a circle, and clockwise means turning in the direction a clock does - right.

So a point would be turned right halfway around.

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What is the least common denominator of 3/4, 4/5, 2/3?

morpeh [17]

D A A C A C A B A D A C B B 1/1/3 B 43/1/5 A A

4 0

3 years ago

Read 2 more answers

Consider the following function. f(x) = 7x + 1 x (a) Find the critical numbers of f. (Enter your answers as a comma-separated li

sukhopar [10]

Answer:

a) DNE

b) The function increases for every real value of x.

c) DNE

Step-by-step explanation:

Given a function f(x), the critical points are those in which $f^{\prime}(x) = 0$ , that is, the roots of the first derivative of f(x).

Those critical points let us find the intervals in which the function increases or decreases. If the first derivative in the interval is positive, the function increases in the interval. If it is negative, the function decreases.

If the function increases before a critical point and then, as it passes the critical point, it starts to decrease, we have that the critical point $(x_{c}, f(x_{c}) is a relative maximum.If the function decreases before a critical point and then, as it passes the critical point, it starts to increase, we have that the critical point [tex](x_{c}, f(x_{c}) is a relative minimum.If the function has no critical points, it either always increases or always decreases.In this exercise, we have that:[tex]f(x) = 7x + 1$

(a) Find the critical numbers of f.

$f^{\prime}(x) = 0$

$f^{/prime}(x) = 7$

7 = 0 is false. This means that f has no critical points.

(b) Find the open intervals on which the function is increasing or decreasing.

Since there are no critical points, we know that either the function increases or decreases in the entire real interval.

We have a first order function in the following format:

$f(x) = ax + b$

In which $a > 0$ .

So the function increases for every real value of x.

(c) Apply the First Derivative Test to identify the relative extremum.

From a), we find that there are no critical numbers. So DNE

7 0

3 years ago

A right rectangular prism is packed with cubes of side length 1/6 inch. If the prism is packed with 12 cubes along the length, 3

Ede4ka [16]

Answer:

a

Step-by-step explanation:

2*1/6 is 2/6, which is 1/3

3 0

3 years ago

Decide whether the equations is a trigonometric function identity, EXPLAIN.

hodyreva [135]

(1) Answer: Identity true

$\cos^2x(1+\tan^2x)=\cos^2x(1+\frac{sin^2x}{cos^2x})=\\= \cos^2x\frac{cos^2x+sin^2x}{cos^2x}=1$

(2) Answer: Identity true

$\sec x \tan x(1-\sin^2x)=\sin x\\\frac{1}{\cos x}\frac{\sin x}{\cos x}(\cos^2x)=\sin x\\\sin x = \sin x$

(3) Answer: Identity true

$\sin^2(\theta)\csc^2(\theta)=\sin^2(\theta)+\cos^2(\theta)\\\sin^2(\theta)\frac{1}{\sin^2(\theta)}=\sin^2(\theta)+\cos^2(\theta)\\1 = 1$

7 0

3 years ago

The midpoint of RS is M(1, 2). One endpoint is R(-6, 6). Find the coordinates of endpoint S.

AysviL [449]

Answer:

The coordinates of point S are (8,-2)

Step-by-step explanation:

we know that

The formula to calculate the midpoint between two points is equal to

$M=(\frac{x1+x2}{2},\frac{y1+y2}{2})$

Let

Point S(x2,y2)

substitute the given values

$(1,2)=(\frac{-6+x2}{2},\frac{6+y2}{2})$

<em>Solve for x2</em>

$1=(-6+x2)/2\\2=-6+x2\\x2=2+6=8$

<em>Solve for y2</em>

$2=(6+y2)/2\\4=6+y2\\y2=4-6=-2$

therefore

The coordinates of point S are (8,-2)

5 0

3 years ago