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3 years ago

11

A car dealer offers 5 different models, 6 colors, and 3 kinds of transmissions. How many unique cars could you choose from at th

is dealership? to this you do 5*6*3 right?

2 answers:

zimovet [89]3 years ago

8 0

The answer is 90 because 5x6x3.

Harlamova29_29 [7]3 years ago

8 0

Answer: 90 A P E X

Step-by-step explanation:

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A student’s work to simplify and combine radicals is shown below. Select the statement which best applies to the sample mathemat

jeka94

Your answer is incorrect. You forgot to get the square root of 25 and 4. Answer should be 16√2

we can only subtract radicals that are the same. At first glance, 4√50 - 2√8 are not the same, so they are not likely to be subtracted. However, each radical can still be simplified.

4 √50 = 4 √25 * 2 = 4 * 5 √2 = 20 √2
2 √8 = 2 √4 * 2 = 2 * 2 √2 = 4 √2

Now that the radicals are the same. then you can subtract the numbers.

20 √2 - 4 √2 = 16√2

4 0

2 years ago

The product of a number y and 3 is 6.<br> Write the word sentence as an equation

Romashka [77]

3y = 6

Product means multiply so 3 multiplied by y is 3y which is equal to 6. :)

4 0

2 years ago

Good morning can y’all help me

Ulleksa [173]

Answer:

rate of change

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

A line in the xy-plane that passes through the coordinate points (3, -6) and (–7, –4) will never intersecta line that is represe

anyanavicka [17]

Answer:

option A

Step-by-step explanation:

given coordinate (3, -6) and (–7, –4)

to find the equation of line

slope of the line passing through both the point will be

$m = \dfrac{y_2-y_1}{x_2-x_1}$

$m = \dfrac{-4+6}{-7-3}$

$m = -\dfrac{2}{10}$

equation of line

( y - y₁ ) = m ( x - x₁ )

$y + 6= -\dfrac{2}{10}(x - 3)$

x + 5 y = -27

line which will not intersect will be parallel to it so option A has same slope as our line equation i.e. -1/5

hence, correct answer is option A

3 0

2 years ago

For waht values of x do the vectors -1,0,-1), (2,1,2), (1,1, x) form a basis for R3?

DaniilM [7]

<h2>Answer:</h2>

The values of x for which the given vectors are basis for R³ is:

$x\neq 1$

<h2>Step-by-step explanation:</h2>

We know that for a set of vectors are linearly independent if the matrix formed by these set of vectors is non-singular i.e. the determinant of the matrix formed by these vectors is non-zero.

We are given three vectors as:

(-1,0,-1), (2,1,2), (1,1, x)

The matrix formed by these vectors is:

$\left[\begin{array}{ccc}-1&2&1\\0&1&1\\-1&2&x\end{array}\right]$

Now, the determinant of this matrix is:

$\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-1(x-2)-2(1)+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+2-2+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+1$

Hence,

$-x+1\neq 0\\\\\\i.e.\\\\\\x\neq 1$

4 0

2 years ago