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Margarita [4]
3 years ago
5

A line passes through the point (-8,8) and has a slope of -3/2. Write an equation in point-slope form for a this line.

Mathematics
1 answer:
zhuklara [117]3 years ago
3 0

Answer:

y-8=-3/2(x+8)

Step-by-step explanation:

y-y1=m(x-x1)

y-8=-3/2(x-(-8))

y-8=-3/2(x+8)

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John volunteers at the animal shelter on weekends. One Saturday, John unloaded 27 bags of dog food. Each bag weighed 35 pounds.
romanna [79]

Answer:

945 ounces

Step-by-step explanation:

if you read it closley you will find that if you multiply you will get 945 ounces because all you have to do is multipply.

5 0
3 years ago
30 Points!!Need Help Please!!! Will give Brainliest!!!
mixer [17]

Since DC = 12, then EF also = 12.

Both AD and BC are indicated to be the same .

So AE = (24 - 12) / 2 = 12/2 = 6


Now you have a right triangle with height DE = 8 and base AE = 6


Using the Pythagorean Theorem we can now find AD:


AD = √(8^2 + 6^2)

AD = √(64 + 36)

AD =√100

AD = 10

3 0
3 years ago
A six-sided fair dle is rolled 4 times in a row. The probability of getting a 4 only on the last trial is ______
4vir4ik [10]

Answer:

1/6

Step-by-step explanation:

No. Of faces contains 4=1

Total no. Of faces=6

Probablity = favourable no. of outcomes/Total no. of outcomes

PLEASE GIVE BRAINLIEST

4 0
3 years ago
A two-factor study with two levels of factor A and three levels of factor B uses a separate group of n = 5 participants in each
Mashcka [7]

Answer: There are 30 participants are needed for the entire study.

Step-by-step explanation:

Since we have given that

Number of levels of factor A = 2

Number of levels of factor B = 3

Number of participants in each treatment condition = 5

So, the number of participants are needed for the entire study is given by

2\times 3\times 5\\\\=6\times 5\\\\=30

Hence, there are 30 participants are needed for the entire study.

7 0
3 years ago
Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.
Anuta_ua [19.1K]

Answer:

c

Step-by-step explanation:

First, we can transform this into a matrix. The x coefficients will be the first ones for each row, the y coefficients the second column, etc.

\left[\begin{array}{cccc}1&-2&3&-2\\6&2&2&-48\\1&4&3&-38\end{array}\right]

Next, we can define a reduced row echelon form matrix as follows:

With the leading entry being the first non zero number in the first row, the leading entry in each row must be 1. Next, there must only be 0s above and below the leading entry. After that, the leading entry of a row must be to the left of the leading entry of the next row. Finally, rows with all zeros should be at the bottom of the matrix.

Because there are 3 rows and we want to solve for 3 variables, making the desired matrix of form

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] for the first three rows and columns. This would make the equation translate to

x= something

y= something

z = something, making it easy to solve for x, y, and z.

Going back to our matrix,

\left[\begin{array}{cccc}1&-2&3&-2\\6&2&2&-48\\1&4&3&-38\end{array}\right] ,

we can start by removing the nonzero values from the first column for rows 2 and 3 to reach the first column of the desired matrix. We can do this by multiplying the first row by -6 and adding it to the second row, as well as multiplying the first row by -1 and adding it to the third row. This results in

\left[\begin{array}{cccc}1&-2&3&-2\\0&14&-16&-36\\0&6&0&-36\end{array}\right]

as our matrix. * Next, we can reach the second column of our desired matrix by first multiplying the second row by (2/14) and adding it to the first row as well as multiplying the second row by (-6/14) and adding it to the third row. This eliminates the nonzero values from all rows in the second column except for the second row. This results in

\left[\begin{array}{cccc}1&0&10/14&-100/14\\0&14&-16&-36\\0&0&96/14&-288/14\end{array}\right]

After that, to reach the desired second column, we can divide the second row by 14, resulting in

\left[\begin{array}{cccc}1&0&10/14&-100/14\\0&1&-16/14&-36/14\\0&0&96/14&-288/14\end{array}\right]

Finally, to remove the zeros from all rows in the third column outside of the third row, we can multiply the third row by (16/96) and adding it to the second row as well as multiplying the third row by (-10/96) and adding it to the first row. This results in

\left[\begin{array}{cccc}1&0&0&-5\\0&1&0&-6\\0&0&96/14&-288/14\end{array}\right]

We can then divide the third row by -96/14 to reach the desired third column, making the reduced row echelon form of the matrix

\left[\begin{array}{cccc}1&0&0&-5\\0&1&0&-6\\0&0&1&-3\end{array}\right]

Therefore,

x=-5

y=-6

z=-3

* we could also switch the second and third rows here to make the process a little simpler

3 0
3 years ago
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