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Nataliya [291]
2 years ago
11

Which table represents a linear functionWhich table represents a linear function

Mathematics
1 answer:
IceJOKER [234]2 years ago
6 0

The table that has a constant average rate of change represents a linear function.

<h3>What is the average rate of change of a function?</h3>

The average rate of change of a function is given by the <u>change in the output of the function divided by the change in the input</u>.

When this rate is constant, the function is said to be linear, hence the table that has a constant average rate of change represents a linear function.

More can be learned about linear functions at brainly.com/question/24808124

#SPJ1

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GEOMETRY-TRIG
Andrej [43]

Answer:

6. The distance Rampy Ronald would have travelled horizontally is approximately 48.428 feet for an actual maximum height of 21.486 feet

7. Rampy Ronald is going to make the jump

Please find attached the graph of the motion of Rampy Ronald, created with Microsoft Excel

Step-by-step explanation:

6. The given parameters are;

The width of the ravine = 80 ft

The inclination of the ramp with which Rampy Ronald will jump = 30°

The distance from the edge of the ravine the ramp ends = 5 feet

The speed with which Rampy Ronald will make the jump = 60 mph

The maximum height he will reach = 30 feet

Therefore, we have;

The height of the end of the ramp = 15 × sin(30°) = 7.5

The height of the end of the ramp = 7.5 feet

The height of the jump = 30 - 7.5 = 22.5

The vertical velocity = 60 × sin(30°) = 30

The vertical velocity = 30 mph

The horizontal velocity = 60 × cos(30°) = 30·√3

The horizontal velocity = 30·√3 mph

From the kinematic equation, v² = u² - 2gh, we have;

u² = 2gh

h = u²/2g = 900/(2 × 32.17405) ≈ 13.986

The maximum height reached 13.986 + 7.5 = 21.486

From v = u - gt

t = u/g = 30/32.17405 ≈ 0.932

t  ≈ 0.932 s

The distance Rampy Ronald would have travelled horizontally = 30·√3 × 0.932 ≈ 48.428

The distance Rampy Ronald would have travelled horizontally ≈ 48.428 feet

7. The time before Rampy Ronald reaches ground level again, is given by the following relation;

h = 1/2·g·t²

21.846 = 1/2 × 32.17405 × t²

t = √(21.846/16.087025) ≈ 1.165

t ≈ 1.366 s

The horizontal distance traveled is therefore;

The horizontal distance traveled from maximum height is 30·√3 × 1.165 ≈ 60.535

The horizontal distance traveled from maximum height is ≈ 70.979 feet

The total horizontal distance traveled ≈ 48.428 + 60.535 = 108.963

The total horizontal distance traveled ≈ 108.963 feet

The total horizontal distance required to make it across the ravine = 5 feet + 80 feet = 85 feet

Therefore, Rampy Ronald is going to make the jump

8 0
2 years ago
Please someone help me please!I'm struggling and I'll give extra point's!
alukav5142 [94]

Answer:

The x variable has an exponent of 2

Step-by-step explanation:

To be a linear equation, the highest power must be 1 on the variables

2x^2 +y =7 is quadratic since x^2 has a power of 2

3 0
2 years ago
Twice a number x, minus 17
LekaFEV [45]

Answer:

2x - 7

Step-by-step explanation:

Twice a number x means x times 2, which can be written as 2x

Minus 17 from 2x

7 0
2 years ago
The town of Lantana needs $14,000 for a new playground Lantana elementary school raised 5358, Lantana Middle School raised $2,83
slavikrds [6]

Answer: The answer is actually 1,676. But you can estimate it to 2,000. so the town of Lantana would have about 2,000 dollasrs left to raise to reach there goal.

Step-by-step explanation:

You would first make sure you add all the number like 5,358+2,834+4,132. And once you get you answer you subtract 14,000 from the number you got!

I really hope that helps.

8 0
2 years ago
Read 2 more answers
PLZ HELP!!! Use limits to evaluate the integral.
Marrrta [24]

Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:

\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]

Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

r_i=\dfrac{2i}n

where 1\le i\le n. Each interval has length \Delta x_i=\frac{2-0}n=\frac2n.

At these sampling points, the function takes on values of

f(r_i)=7{r_i}^3=7\left(\dfrac{2i}n\right)^3=\dfrac{56i^3}{n^3}

We approximate the integral with the Riemann sum:

\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3

Recall that

\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4

so that the sum reduces to

\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}

Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:

\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}

Just to check:

\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28

4 0
2 years ago
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