Which graph represents 2 (cosine (StartFraction 2 pi Over 3 EndFraction) + I sine (StartFraction 2 pi Over 3 EndFraction) ) ?

The diagonal of a rectangle is of length a. It splits each corner forming two angles with a ratio of 1:2. The area of the rectan

HACTEHA [7]

Answer:

Step-by-step explanation:

Given

Length of diagonal is a

Diagonal divides the angle in 1:2

such that $\theta +2\theta =90$ (because angle between two sides is 90)

$3\theta =90$

$\theta =30^{\circ}$

width of rectangle is $b=a\sin \theta =\frac{a}{2}$

Length of rectangle is $L=a\cos 30=\frac{\sqrt{3}}{2}a$

Area of rectangle $A=L\cdot b$

$A=\frac{\sqrt{3}}{2}a\times \frac{a}{2}$

$A=\frac{\sqrt{3}}{4}a^2$

5 0

3 years ago

Find the equation of the line that is perpendicular to y = –3x + 1 and passes though the point (6, 3).

noname [10]

Answer:

A. y=1/3x+1

Step-by-step explanation:

When finding a line that is perpendicular to another line, all you have to do is find the "opposite reciprocal" of the slope. In this case that means writing the slope as the fraction -3/1, and then flipping the fraction over (1/-3) and taking away the negative sign which gets you 1/3 for the slope of your new line. Now, all you have of your new equation is the slope. You need to take your new equation (y=1/3x+b) and plug in the x and y coordinates to that equation, and then solve for the last variable which is b. That solving process goes as follows:

3=1/3*6+b

3=2+b

1=b

now you can replace the b with 1 in your equation to get your final answer of y=1/3x+1

6 0

3 years ago

What set of transformations could be applied to rectangle ABCD to create A″B″C″D″? 'Rectangle formed by ordered pairs A at negat

CaHeK987 [17]

Answer:

Reflection over the y-axis and rotation of 180°

Step-by-step explanation:

We have,

Rectangle ABCD with co-ordinates A(-4,2), B(-4,1), C(-1,1) and D(-1,2).

It is transformed to a new rectangle A'B'C'D' with co-ordinates A'(-4,-2), B'(-4,-1), C'(-1,-1) and D'(-1,-2).

The graph of both the triangles is shown below.

we see that,

The rectangle ABCD is reflected about y-axis and then rotated 180° to obtain A'B'C'D'.

Reflected about y-axis Rotation of 180°

A= (-4,2) (4,2) A'= (-4,-2)

B= (-4,1) (4,1) B'= (-4,-1)

C= (-1,1) (1,1) C'= (-1,-1)

D= (-1,2) (1,2) D'= (-1,-2)

Hence, the second rectangle is formed by: Reflection over the y-axis and rotation of 180°.

5 0

3 years ago

Read 2 more answers

For the function f(x)=|x−7|−4, evaluate f(x+1).

GarryVolchara [31]

Answer:

$\displaystyle{f(x+1)=|x-6|-4}$

Step-by-step explanation:

To evaluate $\displaystyle{f(x+1)}$ , substitute $\displaystyle{x}$ to $\displaystyle{x+1}$ so we should have:

$\displaystyle{f(x+1) = |x+1-7|-4}$

Evaluate the expression and finally, we have:

$\displaystyle{f(x+1)=|x-6|-4}$

Learn More

brainly.com/question/27923727 - How to find inverse and value of function

6 0

2 years ago

Can I get help solving this graph please?

Daniel [21]

see the attached figure with the letters

1) find m(x) in the interval A,B
A (0,100) B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100

2) find m(x) in the interval B,C
B(50,40) C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20

3) find n(x) in the interval A,B
A (0,0) B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x

4) find n(x) in the interval B,C
B(50,60) C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30

5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find <span>h'(x)
</span>h'(x)=-36/25=-1.44

6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x)
h'(x)=18/25=0.72

for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72

<span> h'(x) = 1.44 ------------ > not exist</span>

8 0

3 years ago