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3 years ago

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Write an equation in slope intercept form of a line that is perpendicular to the line y=6x+1 and contains the point (12,-1). Sta

rt with the line in point slope form. Show work! Plz help!

1 answer:

uranmaximum [27]3 years ago

6 0

Answer:

The equation of the line would be y = -1/6x + 1

Step-by-step explanation:

In order to find this, we first need to set the slope. Since the original slope is 6, we know the new slope to be -1/6. This is because perpendicular slopes have opposite and reciprocal slopes. So, we put this and the point into point-slope form.

y - y1 = m(x - x1)

y - -1 = -1/6( x -12)

y + 1 = -1/6x + 2

y = -1/6x + 1

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Step-by-step explanation:

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1. Find the digit that makes 3,71_ divisible by 9.

KiRa [710]

<u>QUESTION 1</u>

We want to find the digit that should fill the blank space to make

$3,71-$

divisible by 9.

If a number is divisible by 9 then the sum of the digits should be a multiple of 9.

The sum of the given digits is,

$3 + 7 + 1 = 11$

Since

$11 + 7 = 18$

which is a multiple of 9.

This means that

$3,717$

is divisible by 9.

The correct answer is B

<u>QUESTION 2</u>

The factors of the number 30 are all the numbers that divides 30 exactly without a remainder.

These numbers are ;

$1,2,3,5,6,10,15,30$

The correct answer is A.

<u>QUESTION 3.</u>

We want to find the prime factorization of the number 168.

The prime numbers that are factors of 168 are

$2,3 \: and \: 7$

We can write 168 as the product of these three prime numbers to obtain,

$168={2}^{3}\times 3\times7$

We can also use the factor tree as shown in the attachment to write the prime factorization of 168 as

$168 ={2}^{3}\times 3\times7$

The correct answer is B.

QUESTION 4.

We want to find the greatest common factor of

$140\:\:and\:\:180$

We need to express each of these numbers as a product of prime factors.

The prime factorization of 140 is

$140={2}^{2}\times 5\times7.$

The prime factorization of 180 is

$180={2}^{2} \times{3}^{2}\times5.$

The greatest common factor is the product of the least degree of each common factor.

$GCF={2}^{2}\times5$

$GCF=20$

The correct answer is A.

QUESTION 5.

We want to find the greatest common factor of

$15,30\: and\:60.$

We need to first find the prime factorization of each number.

The prime factorization of 15 is

$15=3\times5.$

The prime factorization of 30 is

$30=2\times 3\times 5.$

The prime factorization of 60 is

$60={2}^{2}\times3 \times5$

The greatest common factor of these three numbers is the product of the factors with the least degree that is common to them.

$GCF=3 \times5$

$GCF=15$

The correct answer is C.

QUESTION 6

We want to determine which of the given fractions is equivalent to

$\frac{3}{8}.$

We must therefore simplify each option,

$A.\: \: \frac{15}{32}=\frac{15}{32}$

$B.\:\:\frac{12}{32}=\frac{4\times 3}{4\times8}=\frac{3}{8}$

$C.\:\:\:\:\frac{12}{24}=\frac{12\times1}{12\times 2}=\frac{1}{2}$

$D.\:\:\frac{9}{32}=\frac{9}{32}$

The simplification shows that

$\frac{12}{32}\equiv \frac{3}{8}$

The correct answer is B.

QUESTION 7.

We want to express

$\frac{10}{22}$

in the simplest form.

We just have to cancel out common factors as follows.

$\frac{10}{22}=\frac{2\times5}{2 \times11}$

This simplifies to,

$\frac{10}{22}=\frac{5}{11}$

The correct answer is C.

QUESTION 8.

We were given that Justin visited $25$ of the $50$ states.

The question requires that we express $25$ as a fraction of $50.$

This will give us

$\frac{25}{50}=\frac{25\times1}{25\times2}$

We must cancel out the common factors to have our fraction in the simplest form.

$\frac{25}{50}=\frac{1}{2}$

The correct answer is C.

QUESTION 9.

We want to write

$2\frac{5}{8}$

as an improper fraction.

We need to multiply the 2 by the denominator which is 8 and add the product to 5 and then express the result over 8.

This gives us,

$2 \frac{5}{8}=\frac{2\times8+5}{8}$

this implies that,

$2\frac{5}{8}=\frac{16+5}{8}$

$2\frac{5}{8}=\frac{21}{8}$

Sarah needed

$\frac{21}{8}\:\:yards$

The correct answer is D.

QUESTION 10

See attachment

QUESTION 11

We wan to write

$3\: and\:\:\frac{7}{8}$

as an improper fraction.

This implies that,

$3+\frac{7}{8}=3\frac{7}{8}$

To write this as a mixed number, we have,

$3\frac{7}{8}=\frac{3\times8+7}{8}$

This implies that,

$3\frac{7}{8}=\frac{24+7}{8}$

This gives

$3\frac{7}{8}=\frac{31}{8}$

The correct answer is B.

QUESTION 12

We want to find the LCM of $30$ and $46$ using prime factorization.

The prime factorization of 30 is $30=2\times 3\times 5$

The prime factorization of 46 is $40=2\times 23$ .

The LCM is the product of the common factors with the highest degrees. This gives us,

$LCM=2\times \times3 5\times 23$

$LCM=690$

The correct answer is D.

QUESTION 13

We want to find the least common multiple of 3,6 and 7.

The prime factorization of $3$ is $3$ .

The prime factorization of 6 is $6=2\times 3$ .

The prime factorization of 7 is $7$ .

The LCM is the product of the common factors with the highest degrees. This gives us,

$LCM=2\times3 \times7$

$LCM=42$ .

The LCM is 42, therefore 42 days will pass before all three bikes will at the park on the same day again.

The correct answer is B.

See attachment for continuation.

6 0

3 years ago

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