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3 years ago

Help me please I don’t understand what to do

Mathematics

1 answer:

Norma-Jean [14]3 years ago

6 0

Answer:

i dont know please dont hate me

Step-by-step explanation:

<h2>i dont know please dont hate me</h2>

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The table below shows the distance d(t) in feet that an object travels in t seconds: t (seconds) d(t) (feet) 2 64 4 256 6 576 8

Thepotemich [5.8K]

Answer:

128 ft/s; it represents the average speed of the object between 2 seconds and 6 seconds

Step-by-step explanation:

The average rate of change is another way to say slope.

slope = (y2-y1)/(x2-x1)

= (576-64)/(6-2)

= (512/4)

=128 ft/s

This represents the change in ft per second, or the average speed the object is going during the time period.

5 0

3 years ago

Read 2 more answers

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago

Please help ASAP <br> Find the slope of the line graphed below.

irina1246 [14]

Answer:

3/4

Step-by-step explanation:

Hope this helps! :)

8 0

3 years ago

Read 2 more answers

Please help me with number six

Tpy6a [65]

$w = 2x = 2( \frac{y}{3}) \\ v = 2w = 2 \times 2( \frac{y}{3} ) \\ v + w + x + y = 180 \\ \frac{4}{3} y + \frac{2}{3} y + \frac{y}{3} + y = 180 \\ y( \frac{4}{3} + \frac{2}{3} + \frac{1}{3} + 1) = 180 \\ y = \frac{180}{()}$

5 0

3 years ago

How to write 21/6 as a mixed number

NNADVOKAT [17]

Divide using long division. The whole number portion will be the number of times the denominator of the original fraction divides evenly into the numerator of the original fraction, and the fraction portion of the mixed number will be the remainder of the original fraction division over the denominator of the original fraction.

3 and 3 over 6

5 0

3 years ago