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3 years ago

What domain and range are responsible for the function?

Mathematics

1 answer:

faust18 [17]3 years ago

8 0

Answer:

Domain { 0 ≤ r≤ 18}

Range = {0≤ c ≤ 31.5}

Step-by-step explanation:

The domain is the values that r can take

He can goes on each ride up to 1 time

R can be from 0 rides to 18 rides since there are only 18 rides

Domain { 0 ≤ r≤ 18}

The range is the output

If r = 0 then he spends no money

if r =18 then 1.75*18 =31.5

Range = {0≤ C(r) ≤ 31.5}

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13. GEOMETRY The formula for the area of a triangle is A = Loh , where A represents the area, b is the base, and his the height

Setler79 [48]

Answer:

Whoa! You made a rectangle that's twice as big as the triangle! The area of the rectangle is bh=4\times 5 = 20bh=4×5=20b, h, equals, 4, times, 5, equals, 20 square units, so the area of the triangle is \dfrac 12 by = \dfrac 12 \times 4 \times 5 = 10

bh=

×4×5=10start fraction, 1, divided by, 2, end fraction, b, h, equal, start fraction, 1, divided by, 2, end fraction, times, 4, times, 5, equals, 10 square units.

Step-by-step explanation:

8 0

2 years ago

Find the slope of the line passing through the points (-6,6) and (4,-7)

Alla [95]

Answer:

m = -13/10

Step-by-step explanation:

Madison, imagine that you're moving from (-6, 6) to (4, -7) with interest in knowing how much x (the run) and y (the rise) change.

In the first instance, x increases by 10 from -6 and +4. Simultaneously, y decreases by 13 from 6 to -7.

Thus the slope is m = rise / run = m = -13/10

8 0

3 years ago

What is the solution for the system of equations? Use the substitution method to solve. Y=5−3x5x−4y=−3 Enter your answer by fill

Taya2010 [7]

Using the substitution method to solve the linear system the value of x and y is 1 and 2 respectively.

<h3>What is the linear system?</h3>

It is a system of an equation in which the highest power of the variable is always 1. A one-dimension figure that has no width.

Given

The equations are shown below.

y = 5 - 3x .....1

5x - 4y = -3 .....2

Substitute equation 1 in equation 2. Then

$\begin{aligned} 5x - 4(5-3x) &= -3\\\\5x - 20 + 12x &= -3\\\\17x &= -3 +20 \\\\17x &= 17\\\\x &= 1\end{aligned}$

Then substitute the value of x in equation 1.

$y = 5- 3x\\\\y = 5-3*1\\\\y = 2$

Thus, the value of x and y is 1 and 2 respectively.

More about the linear system link is given below.

brainly.com/question/20379472

6 0

1 year ago

Need the answers plz

Shkiper50 [21]

1)
         x^2 + 4x - 5
y =   --------------------
              3x^2 - 12

Vertcial asym => find the x-values for which the equation is not defined and check whether the limit goes to + or - infinite

=> 3x^2 - 12 = 0 => 3x^2 = 12

=> x^2 = 12 / 3 = 4

=> x = +/-2

Limit of y when x -> 2(+) = ( 2^2 + 4(2) - 5) / 0 = - 1 / 0(+) = ∞

Limit of y when x -> 2(-) = - 1 / (0(-) = - ∞

Limit of y when x -> - 2(+) = +∞

Limit of y when x -> - 2(-) = -

=> x = 2 and x = - 2 are a vertical asymptotes

Horizontal asymptote => find whether y tends to a constant value when x -> infinite of negative infinite

Limi of y when x -> - ∞ = 1/3

Lim of y when x -> +∞ = 1/3

=> Horizontal asymptote y = 1/3

x - intercept => y = 0

=>
          x^2 + 4x - 5
0 =   ------------------ => x ^2 + 4x - 5 = 0
              3x^2 - 12

Factor x^2 + 4x - 5 => (x + 5) (x - 1) = 0 => x = - 5 and x = 1

=> x-intercepts x = - 5 and x = 1

Domain: all the real values except x = 2 and x = - 2

2) y = - 2 / (x - 4) - 1

using the same criteria you get:

Vertical asymptote: x = 4

Horizontal asymptote: y = - 1

Domain:all the real values except x = 4

Range: all the real values except y = - 1

3)

x/ (x + 2) + 7 / (x - 5) = 14 / (x^2 - 3x - 10)

factor x^2 - 3x - 10 => (x - 5)(x + 2)

Multiply both sides by (x - 5) (x + 2)

=> x(x - 5) + 7( x + 2) = 14

=> x^2 - 5x + 7x + 14 = 14

=> x^2 + 2x = 0

=> x(x + 2) = 0 => x = 0 and x = - 2 but the function is not defined for x = - 2 so it is not a solution => x = 0

Answer: x = 0

4 0

3 years ago

Is this mapping diagram a function? <br><br>

marshall27 [118]

Include a photo attach a photo of the diagram

8 0

2 years ago