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NARA [144]
3 years ago
11

The equation f = 0 + at represents the final velocity of an object, f, with an initial velocity, v, and an acceleration rate,

Mathematics
1 answer:
balandron [24]3 years ago
4 0

Answer: option 1 is the correct answer

Step-by-step explanation:

This equation is one of the Newton's equations of motion.

The equation is f = v + at which represents the final velocity of an object, f, with an initial velocity, v, and an acceleration rate, a, over time, t.

To determine an equivalent equation solved for a, we will make a the subject of the formula.

f = v+at

subtracting v from both sides of the equation, it becomes

f-v = 0 + at

Dividing both sides of the equation by t, it becomes

f -v = at

a = (f-v)/t

Option 1 is correct answer

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I think the answer is B
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3 years ago
-5/6x = -10/3<br> solve for x
ololo11 [35]

Answer:

x = 4

Step-by-step explanation:

First you need to multiply both sides of the equation by -6/5

-6/5 × (-5/6x) = -6/5 × -10/3

Then you need to calculate and reduce. First, you'll reduce the numbers with the greatest common divisor, 6.

1/5 × 5x = -6/5 × (-10/3)

then reduce the greatest common divisor, 5

x = -6/5 × (-10/3)

Then multiply (multiplying two negatives equals a positive)

x = 6/5 × 10/3

reduce the greatest common divisor, 3

x = 2/5 × 10

reduce the greatest common divisor, 6

x = 2 × 2

x = 4

(i know this is confusing, sorry)

8 0
3 years ago
Graph for f(x)=6^6 and f(x)=14^x
zlopas [31]

Graph Transformations

There are many times when you’ll know very well what the graph of a

particular function looks like, and you’ll want to know what the graph of a

very similar function looks like. In this chapter, we’ll discuss some ways to

draw graphs in these circumstances.

Transformations “after” the original function

Suppose you know what the graph of a function f(x) looks like. Suppose

d 2 R is some number that is greater than 0, and you are asked to graph the

function f(x) + d. The graph of the new function is easy to describe: just

take every point in the graph of f(x), and move it up a distance of d. That

is, if (a, b) is a point in the graph of f(x), then (a, b + d) is a point in the

graph of f(x) + d.

As an explanation for what’s written above: If (a, b) is a point in the graph

of f(x), then that means f(a) = b. Hence, f(a) + d = b + d, which is to say

that (a, b + d) is a point in the graph of f(x) + d.

The chart on the next page describes how to use the graph of f(x) to create

the graph of some similar functions. Throughout the chart, d > 0, c > 1, and

(a, b) is a point in the graph of f(x).

Notice that all of the “new functions” in the chart di↵er from f(x) by some

algebraic manipulation that happens after f plays its part as a function. For

example, first you put x into the function, then f(x) is what comes out. The

function has done its job. Only after f has done its job do you add d to get

the new function f(x) + d. 67Because all of the algebraic transformations occur after the function does

its job, all of the changes to points in the second column of the chart occur

in the second coordinate. Thus, all the changes in the graphs occur in the

vertical measurements of the graph.

New How points in graph of f(x) visual e↵ect

function become points of new graph

f(x) + d (a, b) 7! (a, b + d) shift up by d

f(x) Transformations before and after the original function

As long as there is only one type of operation involved “inside the function”

– either multiplication or addition – and only one type of operation involved

“outside of the function” – either multiplication or addition – you can apply

the rules from the two charts on page 68 and 70 to transform the graph of a

function.

Examples.

• Let’s look at the function • The graph of 2g(3x) is obtained from the graph of g(x) by shrinking

the horizontal coordinate by 1

3, and stretching the vertical coordinate by 2.

(You’d get the same answer here if you reversed the order of the transfor-

mations and stretched vertically by 2 before shrinking horizontally by 1

3. The

order isn’t important.)

74

7:—

(x) 4,

7c’

‘I

II

‘I’

-I

5 0
2 years ago
A researcher is going to estimate the average typing speed of students of a college. He selects a random sample of 20 students a
Stella [2.4K]

Answer:

The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 20 - 1 = 19

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.9}{2} = 0.95. So we have T = 1.7291

The margin of error is:

M = T\frac{s}{\sqrt{n}}

In which s is the standard deviation of the sample and n is the size of the sample. For this question, we have s = 5, n = 20. So

M = T\frac{s}{\sqrt{n}}

M = 1.7291\frac{5}{\sqrt{20}}

M = 1.933

The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.

8 0
3 years ago
This function <br>Is it differentiable at x = 1​ ?
nikklg [1K]

It is not differentiable at x=1 since the slope of the tangent line as x -> 1 from the right is 1 while the slope of the tangent line as x->1 from the left is -1

8 0
1 year ago
Read 2 more answers
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