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EleoNora [17]
3 years ago
10

Help me with this plsss 40 pointsss

Mathematics
1 answer:
lara [203]3 years ago
7 0

Answer:

Step-by-step explanation:

x(3x+2)=16

3x²+2x-16=0

3x²+8x-6x-16=0

x(3x+8)-2(3x+8)=0

(3x+8)(x-2)=0

either 3x+8=0

x=-8/3(rejected as width can't be negative)

x-2=0

x=2

so width=2

length=3×2+2=6+2=8

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Graph for f(x)=6^6 and f(x)=14^x
zlopas [31]

Graph Transformations

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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

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5 0
3 years ago
What’s the length of the mid segment of a trapezoid with bases of length 16 and 24
blagie [28]

Answer:

20

Step-by-step explanation:

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\frac{sumofbases}{2} = \frac{16+24}{2} = \frac{40}{2} = 20

7 0
3 years ago
Solve the following problems.<br> Find x, y
natta225 [31]

Step-by-step explanation:

by using similarity and enlargement

10/8 = 7+y/7

70=56+8y

8y=14

y= 7/4

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7 0
3 years ago
Read 2 more answers
1. (5pts) Find the derivatives of the function using the definition of derivative.
andreyandreev [35.5K]

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

8 0
2 years ago
Describe how the transformed function below is obtained from its parent function. y=x^2 4
Kazeer [188]
I believe its 4 we haven't really went over it
8 0
3 years ago
Read 2 more answers
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