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stealth61 [152]

3 years ago

6

Find the inverse of f(x)= -x + 3

1 answer:

inna [77]3 years ago

4 0

Let g the inverse function of f.

The most important property of g and f being inverses of each other, is that

g(f(x))=x, also f(g(x))=x

so, what one function 'does' to x, the other 'undoes' it.

Thus, we have:

f(g(x))=x and alos f(g(x))= -g(x)+3, from the rule

thus :

-g(x)+3=x

-g(x)=x-3

g(x)=-x+3

check: f(g(x))=f(-x+3)=-(-x+3)+3=x-3+3=x

Answer: the inverse of f is g, such that g(x)=-x+3

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This diagram is a straightedge and compass construction. A is the center of one circle, and B

Serhud [2]

Answer:

Correct choices are a, b, e

Step-by-step explanation:

As per diagram, the circles are equal.

<u>Since the distance AB is the radius of both circles:</u>

AB = AC = BC = BD = AD = r

<u>So the answer choices:</u>

a. AC = BC

Correct, both equal to r

b. AC = BD

Correct, both equal to r

C. CD = AB

Incorrect. Half of CD is the leg of 30-60-90 triangle and is equal to r√3/2, so CD = AB√3

d. ABCD is a square

Incorrect. ACBD is rhombus

e. ABD is an equilateral triangle

Correct, all three sides are equal to r

f. CD = AB + AB

Incorrect as CD = AB√3

3 0

3 years ago

10. Use the diagram below to find the value of x.

Sholpan [36]

Answer:

x=20

Step-by-step explanation:

Hello There!

Remember the exterior angle of a triangle rule:

An exterior angle of a triangle is equal to the sum of the opposite interior angles

Knowing this, we can create an equation to solve for x

exterior angle (100) = sum of opposite interior angles (3x+2x)

100 = 2x+3x

now we solve for x

step 1 combine like terms

2x+3x=5x

now we have 100=5x

step 2 divide each side by 5

5x/5=x

100/5=20

we're left with x = 20

7 0

3 years ago

Read 2 more answers

If limx→3f(x)=7, which of the following must be true? I. f is continuous at x = 3 II. f is differentiable at x = 3

Viktor [21]

Without knowing anything else about $f(x)$ , neither of these need be true.

Suppose

$f(x)=|x-3|+7=\begin{cases}10-x&\text{for }x3\\0&\text{for }x=3\end{cases}$

Then (I) isn't true because, while the limit exists as $x\to3$ and is equal to 7, we have $f(3)=0\neq7$ , so $f(x)$ is not continuous there.

(II) also true because $f(x)$ is not differentiable at $x=3$ ; that is,

$\displaystyle\lim_{x\to3^-}f'(x)=\lim_{x\to3}(10-x)'=\lim_{x\to3}(-1)=-1$

but

$\displaystyle\lim_{x\to3^+}f'(x)=\lim_{x\to3}(x+4)'=\lim_{x\to3}1=1$

which means the derivative does not exist at $x=3$ .

3 0

3 years ago

Can you please help me with this question

user100 [1]

Answer:

tbh im dumb so

Step-by-step explanation:

5 0

3 years ago

1+2<br> 333333+333333333

Rus_ich [418]

335666667........

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5 0

3 years ago