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2 years ago

A line segment can intersect a regular hexagon in ________?

Mathematics

2 answers:

Goryan [66]2 years ago

8 0

Six line segments! So number 3 would be correct.

sattari [20]2 years ago

7 0

The answer is 3. Six points

hope this helps

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What is the answer to #4?

Fudgin [204]

The answer to the question

8 0

3 years ago

Si tus padres te quieren comprar una computadora que cuesta $16.000 pero solo tienen $15.500 y el vendedor les hace un descuento

NNADVOKAT [17]

Answer:

Si podras comprar la computadra con el descuento.

Step-by-step explanation:

?/16 = 6%/100%

16 x 6% = 96%/100% = .96

16 - 0.96 = 15.04

15.50 - 15.04 = 0.46

6 0

2 years ago

Cos(−θ)=√3/3, sinθ<0

Delicious77 [7]

Answer:

$\sin\theta=-\frac{\sqrt{6}}{3}$

Step-by-step explanation:

The given trigonometric equation is $\cos(-\theta)=\frac{\sqrt{3} }{3}$ .

We can either use the Pythagorean identity or the right angle triangle to solve for $\sin\theta$ .

According to the Pythagorean identity,

$\cos^2\theta+\sin^2\theta=1$

Recall that, the cosine function is an even function, therefore

$\cos(-\theta)=\cos(\theta)$

$\Rightarrow \cos(\theta)=\frac{\sqrt{3} }{3}$ .

We substitute this value in to the above Pythagorean identity to get;

$(\frac{\sqrt{3}}{3})^2+\sin^2\theta=1$

$\Rightarrow \frac{3}{9}+\sin^2\theta=1$

$\Rightarrow \sin^2\theta=1-\frac{3}{9}$

$\Rightarrow \sin^2\theta=\frac{6}{9}$

$\Rightarrow \sin\theta=\pm \sqrt{\frac{6}{9}}$

$\Rightarrow \sin\theta=\pm \frac{\sqrt{6}}{3}$

But we were given that,

$\sin\theta\:$ , so we choose the negative value.

$\Rightarrow \sin\theta=-\frac{\sqrt{6}}{3}$

The correct answer is B

7 0

3 years ago

NEED HELP WITH THESE

faltersainse [42]

Step-by-step explanation:

1 .f(x)= $x^{2}$ -x+1

f(-1)= (-1)-(-1)+1

f(-1)= 1+1+1

= 3

3.f(x)= $x^{2}$ -x+1

f( 1)=(1)-1+1

f( 1)=1-1+1

5. f(x)= $x^{2}$ -x+1

f(3)= $(3)^{2}$ -3+1

= 9-3+1

2. g(x) = 5 - 3x

g( -8)=5-3(-8)

g( -8)=5+24

=29

4.g(x) = 5 - 3x

g(5)=5-3(5)

g(5)=5-15

=-10

6.g(x) = 5 - 3x

g(-3)=5-3(-3)

g(-3)=5+9

=14

3 0

2 years ago

Consider the two lines l1:x=−2t,y=1+2t,z=3tl1:x=−2t,y=1+2t,z=3t and l2:x=−8+4s,y=0+5s,z=5+1sl2:x=−8+4s,y=0+5s,z=5+1s find the po

seraphim [82]

At the point of intersection, the coordinates are the same:

... (x, y, z) = (-2t, 1+2t, 3t) = (-8+4s, 5s, 5+s)

We only need two of the coordinates to solve for the values of s and t. Using the x- and y-coordinates, we have

... 4s + 2t = 8 . . . . . . x2 - x1 = 0, in standard form

... 5s -2t = 1 . . . . . . . y2 - y1 = 0, in standard form

Adding these equations gives 9s=9, so s=1. Substituting into either equation gives t=2.

Using the expression for l1 with t=2, the point of intersection is

... (-2·2, 1+2·2, 3·2) = (-4, 5, 6).

_____

We could have stopped after finding the value of s, because that defines the point. By finding the value of t, we can check the solution to make sure that l1 and l2 both give the same point for the respective values of s and t. (They do.)

8 0

3 years ago