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3 years ago

How do u do this with answer plz

Mathematics

2 answers:

kap26 [50]2 years ago

7 0

You have to find the area
First find the area of the triangle and then of the rectangle by finding the length and the with of each side of the rectangle multiplying it and than adding them

Whitepunk [10]2 years ago

6 0

Answer:

i guess u know how to calculate a rectangle

Step-by-step explanation:

https://sciencing.com/area-triangular-prism-8165114.html

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In a sample of 100 randomly selected concert attendees, the concert organizers find that 11 of them left the show early. The con

3241004551 [841]

Answer:

A) ≈22.4 concert attendees might have left early.

B) 605 attendees might leave early.

Step-by-step explanation:

A) 11/100, or 11%, of the sample left early.

11% of all the concert attendees is (11/100) × 2040 ≅ 22.4

attendees who might have left early.

B) 11% of the total attendance is 11% × 5500 = 605 attendees who might

leave early.

8 0

3 years ago

Julian counted the number of times he drove across the Seven Mile Bridge while vacationing in the Florida keys. He crossed the b

timofeeve [1]

238 Miles. a 7 mile bridge is 7 miles and he drove across it 34 times so 7*34

7 0

3 years ago

Read 2 more answers

Pls help this is pretty urgent

laiz [17]

Answer:

(a) See below.

(b) x = 0 or x = 1

Step-by-step explanation:

Given rational function:

$f(x)=\dfrac{\ln |x-1|}{x}$

Part (a)

Substitute x = 2 into the given rational function:

$\begin{aligned}\implies f(2) & =\dfrac{\ln |2-1|}{2}\\\\ & =\dfrac{\ln 1}{2}\\\\ & =\dfrac{0}{2}\\\\ & =0\end{aligned}$

Therefore, as the function is defined at x = 2, the function is continuous at x = 2.

Part (b)

Given interval: [-2, 2]

Logs of negative numbers or zero are undefined. As the numerator is the natural log of an absolute value, the numerator is undefined when:

|x - 1| = 0 ⇒ x = 1.

A rational function is undefined when the denominator is equal to zero, so the function f(x) is undefined when x = 0.

So the function is discontinuous at x = 0 or x = 1 on the interval [-2, 2].

Part (c)

x = 1 is a vertical asymptote. As the function exists on both sides of this vertical asymptote, it is an infinite discontinuity. Since the function doesn't approach a particular finite value, the limit does not exist. Therefore, x = 1 is a non-removable discontinuity.

A hole exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal.

$\implies f(0)=\dfrac{\ln |0-1|}{0}=\dfrac{\ln 1}{0}=\dfrac{0}{0}$

Therefore, there is a hole at x = 0.

The removable discontinuity of a function occurs at a point where the graph of a function has a hole in it. Therefore, x = 0 is a removable discontinuity.