What kind of shape is this ?

Triangle PQR and triangle QRS have vertices P(-9,7) , Q(4,7) , R(4,-3), and S(-10,3) What is the area in square units of the qua

larisa [96]

Answer:

3.7

Step-by-step explanation:

.

7 0

3 years ago

Jordie is getting balloons for his aunt's birthday party. He wants each balloon string to be 12 feet long. At the party store, s

geniusboy [140]

Answer:

216 yards

Step-by-step explanation:

You'd multiply the amount of string for each balloon (12 feet) by the 54 balloons he needs, and you get 648 feet. There are 3 feet in a yard so then you would divide 648 by 3 and get 216 yards

8 0

2 years ago

Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is incre

Ainat [17]

Answer: (a) Interval where f is increasing: (0.78,+∞);

Interval where f is decreasing: (0,0.78);

(b) Local minimum: (0.78, - 0.09)

(c) Inflection point: (0.56,-0.06)

Interval concave up: (0.56,+∞)

Interval concave down: (0,0.56)

Step-by-step explanation:

(a) To determine the interval where function f is increasing or decreasing, first derive the function:

f'(x) = $\frac{d}{dx}$ [ $x^{4}ln(x)$ ]

Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),

you have:

f'(x) = $4x^{3}ln(x) + x_{4}.\frac{1}{x}$

f'(x) = $4x^{3}ln(x) + x^{3}$

f'(x) = $x^{3}[4ln(x) + 1]$

Now, find the critical points: f'(x) = 0

$x^{3}[4ln(x) + 1]$ = 0

$x^{3} = 0$

x = 0

and

$4ln(x) + 1 = 0$

$ln(x) = \frac{-1}{4}$

x = $e^{\frac{-1}{4} }$

x = 0.78

To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:

interval x-value f'(x) result

0<x<0.78 0.5 f'(0.5) = -0.22 decreasing

x>0.78 1 f'(1) = 1 increasing

With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.

Note: As it is a natural logarithm function, there are no negative x-values.

(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:

Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
After 0.78, it increase (has a change of sign) and f is also defined;

Then, x=0.78 is a point of minimum and its y-value is:

f(x) = $x^{4}ln(x)$

f(0.78) = $0.78^{4}ln(0.78)$

f(0.78) = - 0.092

The point of minimum is (0.78, - 0.092)

(c) To determine the inflection point (IP), calculate the second derivative of the function and solve for x:

f"(x) = $\frac{d^{2}}{dx^{2}}$ [ $x^{3}[4ln(x) + 1]$ ]

f"(x) = $3x^{2}[4ln(x) + 1] + 4x^{2}$

f"(x) = $x^{2}[12ln(x) + 7]$

$x^{2}[12ln(x) + 7]$ = 0

$x^{2} = 0\\x = 0$

and

$12ln(x) + 7 = 0\\ln(x) = \frac{-7}{12} \\x = e^{\frac{-7}{12} }\\x = 0.56$

Substituing x in the function:

f(x) = $x^{4}ln(x)$

f(0.56) = $0.56^{4} ln(0.56)$

f(0.56) = - 0.06

The inflection point will be: (0.56, - 0.06)

In a function, the concave is down when f"(x) < 0 and up when f"(x) > 0, adn knowing that the critical points for that derivative are 0 and 0.56:

f"(x) = $x^{2}[12ln(x) + 7]$

f"(0.1) = $0.1^{2}[12ln(0.1)+7]$

f"(0.1) = - 0.21, i.e. Concave is DOWN.

f"(0.7) = $0.7^{2}[12ln(0.7)+7]$

f"(0.7) = + 1.33, i.e. Concave is UP.

4 0

3 years ago

Evaluate each expression for g = -7 and h = 3 and match it to its value.

Arturiano [62]

Answer:

The values of given expressions are:

1. gh = -21

2. g^2 - h = 46

3. g + h^2 = 2

4. g + h = -4

5. h - g = 10

6. g - h = -10

Step-by-step explanation:

Given values of g and h are:

g = -7

h = 3

1. gh

The two numbers are being multiplied

Putting the values

$gh = (-7)(3) = -21$

2. g^2-h

Putting the values

$=(-7)^2-3\\=49-3\\=46$

3. g+h^2

Putting the values

$= -7 + (3)^2\\=-7+9\\=2$

4. g+h

Putting the values

$= -7+3\\=-4$

5. h-g

Putting the values

$= 3 - (-7)\\=3+7\\=10$

6. g-h

Putting values

$=-7-3\\=-10$

Hence,

The values of given expressions are:

1. gh = -21

2. g^2 - h = 46

3. g + h^2 = 2

4. g + h = -4

5. h - g = 10

6. g - h = -10

4 0

3 years ago

Read 2 more answers

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

3 years ago