The equation of the graphed line in point-slope form is A , and its equation in slope-intercept form is B .

Four less than the quotient of a number 2 and 5.

Papessa [141]

Answer:

2/5 - 4

Step-by-step explanation:

quotient means to divide

4 less means subtraction

8 0

3 years ago

The two functions below are used to represent the height of a rocket above the ground after it is launched. Which rocket reaches

Korvikt [17]

Answer:

Rocket B reaches a maximum height that is greater than the maximum height of Rocket A.

Step-by-step explanation:

A graph quickly shows the answer to this question.

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In vertex form, the equation for Rocket A is ...

y = -6(x² -16x +8²) +6(8²)

y = -6(x -8)² +384 . . . . . . . . . maximum height is 384 feet

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The corresponding equation for Rocket B is ...

y = -4(x² -20x +10²) +4(10²)

y = -4(x -10)² +400 . . . . . . . . maximum height is 400 feet

Rocket B reaches a maximum height greater than that of Rocket A.

7 0

3 years ago

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If a line falls on the points (6,22) and (4,21), what is its slope? Enter your answer as a fraction in lowest terms. Use a slash

I am Lyosha [343]

Answer:

1/2

Step-by-step explanation:

Since we have two points, we can use the slope formula

m = (y2-y1)/(x2-x1)

= (22-21)/(6-4)

= 1/2

8 0

3 years ago

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Is (4,9) a solution to the equation 3(x+2) = y ?

Mrac [35]

Answer:

(4, 9) is not a solution to 3(x + 2) = y.

Step-by-step explanation:

3(x + 2) = y

Substitute (4, 9) into the equation.

3(4 + 2) = 9

Add the inside of the parenthesis.

3(6) = 9

Multiply.

18 ≠ 9

8 0

3 years ago

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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