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2 years ago

6

Is this graph continuous at x=1?

1 answer:

drek231 [11]2 years ago

8 0

Answer:

Step-by-step explanation:

By my definition, no. There's a continuity up to and including just before 1 approaching from the left, and there's a continuity just after 1 going to the right. But 1 itself has a special definition h(x) = 4 when x = 1 and that makes it discontinuous.

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anygoal [31]

8 ÷ (7 - 9) * ( 4 + (-4) ) <--- notice that bolded part.

4 + (-4) = 4 - 4 = 0.

after that, you're pretty much dividing and multiplying by 0, so

whatever ÷ whatever * whatever * 0 = 0.

6 0

3 years ago

Find the length of the leg(s) of a right triangle, to the nearest hundredth, if both legs are equal and the hypotenuse is 10.

charle [14.2K]

Answer:

7.07 units

Step-by-step explanation:

Use the Pythagorean theorem for the right triangle:

$\text{hypotenuse}^2=\text{leg}_1^2+\text{leg}_2^2,\\ \\c^2=a^2+b^2.$

If a = x, b = x, and c = 10, then

$10^2=x^2+x^2,\\ \\100=2x^2,\\ \\x^2=50,\\ \\x=\sqrt{50}=5\sqrt{2}\approx 7.07\ units.$

7 0

2 years ago

Lol not good with averages pls help

Ivenika [448]

15 is the answer i divided

6 0

2 years ago

Read 2 more answers

A) Find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s.

Fed [463]

Answer:

A) $a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}$

B) $a_{0} = a_{1} = 0$

C) for n = 2

$a_{2}$ = 1

for n = 3

$a_{3}$ = 3

for n = 4

$a_{4}$ = 8

for n = 5

$a_{5}$ = 19

Step-by-step explanation:

A) A recurrence relation for the number of bit strings of length n that contain a pair of consecutive Os can be represented below

if a string (n ) ends with 00 for n-2 positions there are a pair of consecutive Os therefore there will be : $2^{n-2}$ strings

therefore for n ≥ 2

The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os

$a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}$

b ) The initial conditions

The initial conditions are : $a_{0} = a_{1} = 0$

C) The number of bit strings of length seven containing two consecutive 0s

here we apply the re occurrence relation and the initial conditions

$a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}$

for n = 2

$a_{2}$ = 1

for n = 3

$a_{3}$ = 3

for n = 4

$a_{4}$ = 8

for n = 5

$a_{5}$ = 19

7 0

2 years ago

Please help me answer this !! <br> Explain your answer !!! <br><br> I will give brainlst

Ahat [919]

The answer would be 18

8 0

2 years ago