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3 years ago

Plz help ASAP !!!!!!!!!

Mathematics

2 answers:

SashulF [63]3 years ago

5 0

Answer:

45 degrees

Step-by-step explanation:

As the two angles are on a line, they are supplementary. Therefore, x = 180 - 135.

Nana76 [90]3 years ago

5 0

Answer:

47 degrees.

Step-by-step explanation:

I believe 47 degrees is the correct answer because the angle of the entire thing is 180. So take 180 - 133 and you should get 47. I believe this is the correct answer. Hope this helped!

-Kirito

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Tickets to a movie cost $7.25 for adults and $5.50 for students a group of friends purchased 8 tickets for $52.75 how many adult

irina1246 [14]

Answer:

The number of adults tickets and student tickets purchased is 5 and 3.

Given that,

Tickets to a movie cost $7.25 for adults and $5.50 for students.

A group of friends purchased 8 tickets for $52.75.

Here we assume the no of adult tickets and no of student tickets be x and y.

Based on the above information, the calculation is as follows:

7.25x + 5.50y = 52.75 .............(i)

x + y = 8

x = 8 - y..................(2)

Now put the x value in the equation (1)

So,

7.25(8-y) + 5.50y = 52.75

7.25 × 8 - 7.25y + 5.50y = 52.75

58 - 1.75y = 52.75

5.25 = 1.75y

y = 3

So,

x = 8 - 3

= 5

Therefore we can conclude that the number of adults tickets and student tickets purchased is 5 and 3.

-Hope this helps<3

4 0

2 years ago

1) Find x, if 1/3 of x is 14

BartSMP [9]

"of" means multiply and "is" means equal

1. $\frac{1}{3}$ x = 14

x = 14(3) multiply both sides by 3 to isolate "x"

x = 42

2. $\frac{1}{5}$ x = 14

x = 14(5) multiply both sides by 5 to isolate "x"

x = 70

3. $\frac{2}{5}$ x = 14

2x = 14(5) multiply both sides by 5

2x = 70

x = 35 divide both sides by 2 to isolate "x"

4. $\frac{7}{8}$ x = 14

7x = 14(8) multiply both sides by 8

7x = 112

x = 16 divide both sides by 7 to isolate "x"

Answers: 42, 70, 35, 16

5 0

3 years ago

Help Me Please I will Give Brainliest If Correct!!!

disa [49]

Answer:

8^-2a^7 = a^7/8^2 = a^7/64

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Find the line through (3, 1, −2) that intersects and is perpendicular to the line x = −1 + t, y = −2 + t, z = −1 + t. (HINT: If

mel-nik [20]

Answer:

( xo , yo , zo ) = ( 1 , 0 , 1 )

Step-by-step explanation:

Given:-

- A line passing through point (3, 1, −2) intersects and is perpendicular to line with coordinates:

x = −1 + t, y = −2 + t, z = −1 + t .... t = arbitrary parameter.

Find:-

The coordinates for point of intersection.

Solution:-

- The line that passes through point (3, 1, −2) = ( a, b , c ) and an a arbitrary point on the given line have the following direction vector d2 :

d2 = ( x2 , y2 , z2 )

x2 = a - ( x ) = 3 - ( -1 + t ) = 4 - t

y2 = b - ( y ) = 1 - ( -2 + t ) = 3 - t

z2 = c - ( z ) = -2 - ( -1 + t ) = -1 - t

d2 = ( 4 - t , 3 - t , -1 - t )

- The direction vector d1 of the given line is:

d1 = ( x1 , y1 , z1 )

x1 = 1

y1 = 1

z1 = 1

d1 = ( 1 , 1 , 1 )

- The dot product of two orthogonal vectors is always equal to zero:

d1.d2 = 0

( 4 - t , 3 - t , -1 - t ) . ( 1 , 1 , 1 ) = 0

- Solve for parameter (t):

(4 - t) + (3 - t) + (-1 - t) = 0

6 -3t = 0

t = 2

- The coordinates of the point of intersections can be evaluated by substituting the value of "t" into the given equation of line:

xo ( t = 2) = - 1 + 2 = 1

yo ( t = 2) = - 2 + 2 = 0

zo ( t = 2) = - 1 + 2 = 1

- The coordinates are:

( xo , yo , zo ) = ( 1 , 0 , 1 )

8 0

3 years ago

Find the equation of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit.

leonid [27]

Answer-

The equations of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit are

$12x-5y+14=0 \\ 12x-5y-14=0$

Solution-

Let a point which is 1 unit away from the line 12x-5y-1=0 is (h, k)

The applying the distance formula,

$\Rightarrow \left | \frac{12h-5k-1}{\sqrt{12^2+5^2}} \right |=1$

$\Rightarrow \left | \frac{12h-5k-1}{\sqrt{169}} \right |=1$

$\Rightarrow \left | \frac{12h-5k-1}{13} \right |=1$

$\Rightarrow 12h-5k-1=\pm 13$

$\Rightarrow 12h-5k=\pm 14$

$\Rightarrow 12h-5k=14,\ 12h-5k=-14$

$\Rightarrow 12h-5k-14=0,\ 12h-5k+14=0$

$\Rightarrow 12x-5y-14=0,\ 12x-5y+14=0$

Two equations are formed because one will be upper from the the given line and other will be below it.

4 0

3 years ago