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3 years ago

Write the equation of the line y=(5)/(6)x+(7)/(6) perpendicular to

Mathematics

1 answer:

Travka [436]3 years ago

4 0

Answer:

6x + 5y + 13 = 0.

Step-by-step explanation:

y = 5/6x + 7/6

Gradient = 5/6

Since the line is perpendicular to y = 5/6x + 7/6

then its gradient is -6/5.

Hence its equation is: point (-8,7).

y - 7 = -6/5(x -(-8))

multiplying through by 5 we get;

5y - 35 = -6(x + 8)

5y - 35 = -6x - 48

6x + 5y + 13 = 0

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Pls help ASAP I DONT have time

Law Incorporation [45]

Download Symbolab it will answer your question super quick in 2 seconds

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3 years ago

Aaron wants to mulch his garden. His garden is x^2+18x+81 ft^2 One bag of mulch covers x^2-81 ft^2 . Divide the expressions and

blsea [12.9K]

Answer:

Step-by-step explanation:

Given

Garden: $x^2+18x+81$

One Bag: $x^2 - 81$

Requires

Determine the number of bags to cover the whole garden

This is calculated as thus;

$Bags = \frac{x^2+18x+81}{x^2 - 81}$

Expand the numerator

$Bags = \frac{x^2+9x+9x+81}{x^2 - 81}$

$Bags = \frac{x(x+9)+9(x+9)}{x^2 - 81}$

$Bags = \frac{(x+9)(x+9)}{x^2 - 81}$

Express 81 as 9²

$Bags = \frac{(x+9)(x+9)}{x^2 - 9\²}$

Evaluate as difference of two squares

$Bags = \frac{(x+9)(x+9)}{(x - 9)(x+9)}$

$Bags = \frac{(x+9)}{(x - 9)}$

Hence, the number of bags is $Bags = \frac{(x+9)}{(x - 9)}$

3 0

3 years ago

A cyclist travels 3km in 10min what's his speed in km/h

denis-greek [22]

10minutes= 10/60 hours
= 1/6 hrs
speed = distance/ time
=3/ (1/6)
= 18 km/h

5 0

3 years ago

Read 2 more answers

The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,425.

snow_lady [41]

A suitable probability app shows the 1st percentile to be at about 11040 pages.

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The app shown has an error in the first decimal digit. The number is closer to 11040.82301. We expect the manufacturer would round to the nearest 10 or 50 or 100 in advertising.

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3 years ago

A fair coin is tossed three times and the events A, B, and C are defined as follows: A: \{ At least one head is observed \} B: \

Yanka [14]

Answer:

a) $P(A)=0.875$

b) $\text{P(A or B)}=0.875$

c) $\text{P((not A) or B or (not C))}=0.625$

Step-by-step explanation:

Given : A fair coin is tossed three times and the events A, B, and C are defined as follows: A: At least one head is observed, B: At least two heads are observed, C: The number of heads observed is odd.

To find : The following probabilities by summing the probabilities of the appropriate sample points ?

Solution :

The sample space is

S={HHH,HHT,HTT,HTH,TTT,TTH,THH,THT}

n(S)=8

A: At least one head is observed

i.e. A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

n(A)=7

B: At least two heads are observed

i.e. B={HHH,HTT,TTH,THT}

n(B)=4

C: The number of heads observed is odd.

i.e. C={HHH,HTT,THT,TTH}

n(c)=4

a) Probability of A, P(A)

$P(A)=\frac{n(A)}{n(S)}$

$P(A)=\frac{7}{8}$

$P(A)=0.875$

b) P(A or B)

Using formula,

$\text{P(A or B)}=P(A)+P(B)-\text{P(A and B)}$

$\text{P(A or B)}=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{\text{n(A and B)}}{n(S)}$

$\text{P(A or B)}=\frac{7}{8}+\frac{4}{8}-\frac{4}{8}$

$\text{P(A or B)}=\frac{7}{8}$

$\text{P(A or B)}=0.875$

(c) P((not A) or B or (not C))

A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

not A = {TTT} = 1

B={HHH,HTT,TTH,THT}

C={HHH,HTT,THT,TTH}

not C = {HHT,HTH,THH,TTT} = 4

So, not A or B or not C = {HHH,HHT,HTH,THH,TTT}=5

$\text{P((not A) or B or (not C))}=\frac{5}{8}$

$\text{P((not A) or B or (not C))}=0.625$

4 0

3 years ago