Line KL has an equation of a line y = 4x + 5. Which of the following could be an equation for a line that is perpendicular to line AB? (6 points)y = 4x − 8

y = 1 over 4x − 8

y = −4x − 8

y = −1 over 4x − 8

Answers

Answer 1
Answer: To find a perpendicular slope (or line), the slope (in this case 4x) must be the opposite sign and its reciprocal, which is basically the fraction flipped upside down. Since 4 is technically 4/1, that fraction flipped is 1/4. And since you need to flip the sign too, instead of it being a positive number, it's negative. Your answer is -1/4x-8
Answer 2
Answer: y = -1 over 4x - 8 is the answer I think

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The graphs will both have their y-Intercept equal to 8.

Answers

Answer:

d

Step-by-step explanation:

Answer:

8

Step-by-step explanation:

(01.03) A box has dimensions of 17 inches long, 1.3 feet wide, and 8 inches high. What is the volume of the box? The formula for the volume is V = l ⋅ w ⋅ h. (1 point) 2121.6 cubic inches (in3) 176.8 cubic inches (in3) 14.7 cubic inches (in3) 1.2 cubic inches (in3)

Answers

we know that

the volume of a rectangular prism is equal to

V=L*W*H

where

L is the length of the box

W is the width of the box

H is is the height of the box

in this problem we have

L=17\ in  

W=1.3\ ft

H=8\ in

Convert 1.3 feet to inches

we know that

1 feet is equal to 12 inches

so

1.3*12=15.6\ inches

W=15.6 in

Find the volume

V=17*15.6*8

V=2,121.6\ in^(3)

therefore

the answer is

2,121.6\ in^(3)

Convert 1.3 ft to inches which is 15.6in

then multiply all measurements using the volume formula which is L•W•H

17 * 15.6 * 8 = 2121.6 cubic inches

Answer and explanation please

Answers

Answer:

\sf log 162 = p + 4q

Step-by-step explanation:

Given:

  • p = log 2
  • q = log 3

To find :

  • log 162 in terms of p and q.

Solution:

In order to find the logarithm of 162 in terms of p and q, we can use the properties of logarithms.

We can start by expressing 162 as a product of prime factors:

\sf 162 = 2 * 3 * 3 * 3 * 3

Now, we can use the properties of logarithms to simplify this expression:

\sf log 162 = log (2 * 3 * 3 * 3 * 3)

Since log(ab) = log(a) + log(b), we can split this into separate logarithms:

\sf log 162 = log 2 + log (3 * 3 * 3 * 3)

Now, we can use the fact that q = log 3:

\sf log 162 = log 2 + log (3^4)

Using the property\sf \boxed{\sf log(a^b) = b * log(a)}, we get:

\sf log 162 = log 2 + 4 log 3

Now, substitute the values of p and q:

\sf log 162 = p + 4q

So, the logarithm of 162 in termsof p and q is:

\sf log 162 = p + 4q

Answer:

log 162 = 6p + 2q

Step-by-step explanation:

To write log 162 in terms of p and q, we can use the following steps:

- First, we can write 162 as a product of powers of 2 and 3, such as 162 = 2 x 3^4.

- Next, we can use the property of logarithms that log ab = log a + log b to write log 162 = log 2 + log 3^4.

- Then, we can use another property of logarithms that log a^n = n log a to write log 3^4 = 4 log 3.

- Finally, we can substitute p = log 2 and q = log 3 to get log 162 = p + 4q.

We can write 162 as follows:

```

162 = 2^6 * 3^2

```

Therefore,

```

log 162 = log (2^6 * 3^2)

```

Using the logarithmic properties of addition and multiplication, we can simplify this to:

```

log 162 = 6 * log 2 + 2 * log 3

```

Finally, substituting p = log 2 and q = log 3, we get the following expression:

```

log 162 = 6p + 2q

```

Therefore, log 162 can be written as **6p + 2q** in terms of p and q.

Okay, let's break this down step-by-step:

* log 162 = log (2^4 * 3^2)   (by prime factorization)

* log (2^4 * 3^2) = 4log2 + 2log3  (by properties of logarithms)  

* Let p = log 2 and q = log 3

* Substituting:

* log 162 = 4p + 2q

Therefore, log 162 can be written as 4p + 2q, where p = log 2 and q = log 3.

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To express log 162 in terms of p (log 2) and q (log 3), you can use logarithm properties, particularly the change of base formula. The change of base formula states that:

log_b(a) = log_c(a) / log_c(b)

In your case, you want to find log 162:

log 162 = log 2^1 * 3^4

Now, we can use the change of base formula with base 10 (or any other base):

log 162 = (log 2^1 * 3^4) / (log 10)

Since log 10 is simply 1 (logarithm of 10 to any base is 1), we can simplify further:

log 162 = (log 2^1 * 3^4) / 1

Now, apply the properties of logarithms to split the logarithm of a product into a sum of logarithms:

log 162 = (log 2^1) + (log 3^4)

Now, we can replace log 2 with p and log 3 with q:

log 162 = p + (4q)

So, log 162 in terms of p and q is:

log 162 = p + 4q

To write log 162 in terms of p and q, we can use the following steps:

- First, we can write 162 as a product of powers of 2 and 3, such as 162 = 2 x 3^4.

- Next, we can use the property of logarithms that log ab = log a + log b to write log 162 = log 2 + log 3^4.

- Then, we can use another property of logarithms that log a^n = n log a to write log 3^4 = 4 log 3.

- Finally, we can substitute p = log 2 and q = log 3 to get log 162 = p + 4q.

Write 3/13 as a recurring decimal. HELP PLEASE !!!

Answers

Answer:

the answer is 0.23076923076923

Answer: 0.230769 repeat

What are the coordinates of the y-intercept of the line whose equation is LaTeX: 12x+13y=812 x + 13 y = 8? ( , )

Answers

The coordinates of the y-intercept of the line whose equation is

12 x + 13 y = 8 is 8/13.

As given in the question,

Given equation: 12x + 13 y=8

Convert the equation into y-intercept form

General form of y-intercept form is

y=mx + b

Subtract from the equation 12x from both the side of equation,

12x+13y-12x=-12x+8

⇒ 13y=-12x +8

Divide both the side by 13

13y/13= (-12/13)x +8/13

⇒y=(-12/13)x +8/13

To get y-intercept put x=0

y =8/13

Therefore, thecoordinates of the y-intercept of the line whose equation is 12 x + 13 y = 8 is 8/13.

The complete question is :

What are the coordinates of the y-intercept of the line whose equation is

12 x + 13 y = 8 ?

Learn more about y- intercept here

brainly.com/question/14180189

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For the rational equation,(x^(2)+5x+6)/(x+3)=1 , what is a valid value of x?

Answers

Answer:

-3 or -1 is a valid value for x

Step-by-step explanation:

We start by cross multiplying;

So the expression becomes;

x^2 + 5x + 6 = 1(x + 3)

x^2 + 5x + 6 = x + 3

x^2 + 5x -x + 6-3 = 0

x^2 + 4x + 3 = 0

x^2 + x + 3x + 3 = 0

x(x + 1) + 3(x + 1) = 0

(x + 3)(x + 1) = 0

x + 3 = 0 or x + 1 = 0

x = -3 or x = -1