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Katen [24]
3 years ago
7

What is the slope of a line perpendicular to the line y = 1/5x - 3

Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
4 0

Answer:

-5

Step-by-step explanation:

Perpendicular lines have negative reciprocal slopes.

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Round each number to the nearest hundred and estimate the sum.
murzikaleks [220]
1841 ≈ 1800 (nearest hundred)
964 ≈ 1000 (nearest hundred}

<span>1,841 + 964 </span>≈ 1800 + 1000 = 2800 (Answer B)
8 0
3 years ago
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Need help finding the value of x and y
kap26 [50]

Answer:

The values of x = 12 and y = 8.

Step-by-step explanation:

From the given figure ,

ΔMTW≅ΔBGK

That is, these two triangles are congruent.

If two triangles are congruent , all the corresponding angles and corresponding sides are equal.

Congruency is different from similarity . Similarity means two triangles which are the same with different dimensions.

Therefore , ∠MTW = ∠BGK

                     (4x - 3)° = 45°

                       4x = 48°

                         x = 12

Since ∠MTW = 45° ,

∠TMW = 180 - (41 +45)

              = 180 - 86

              =94°

From congruency ,

∠TMW = ∠GBK

94° = 11y + 6

11y = 88°

y = 8

4 0
3 years ago
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
3 years ago
What do you round 150 to??
Vedmedyk [2.9K]
It depends on what you are rounding off to.

If you are round it off to the nearest ten, you must look at the unit number.
If you are rounding off to the nearest hundred, you must look at the ten number.

In this case I think they are asking to round to the nearest hundred. Now we must look at the ten number, which is the number after the 'hundred' number.

The ten number is '5' and the hundred number is '1'. If the ten number is 5 or above, it changes to 0 and it makes the hundred number one higher.

So because the ten number is 5, it changes to 0 and it makes the hundred number one higher, to become 2.

The number is now 200.
5 0
3 years ago
The sum of a number and two times a smaller number is 62. Three times the bigger number exceeds the smaller number by 116. The b
Musya8 [376]
The bigger number is 42 and the smaller number is 10
8 0
3 years ago
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