All categories

3 years ago

Which of the following lines has a negative slope?

Mathematics

1 answer:

Ann [662]3 years ago

3 0

Answer: 3rd graph

Explanation: In order to determine which graph shows a downward or negative slope, we need to be able to look at each of the graphs and determine which line is decreasing when going from left to right.

Notice that the 3rd graph is the only graph with a downward

or negative slope which means that this is our answer.

You might be interested in

7. If P = [5,6,7,8). Q = (2,4,6,8) and<br>R=2, 3, 7, 9), then P nQnR=

Setler [38]

Answer:

pnQnR= {}

Step-by-step explanation:

there is no number common in set P, set Q and set R.

So the answer is null

{}

7 0

3 years ago

Read 2 more answers

Find an equation of variation where y varies directly as x and y = 0.8 when x = 0.4

GrogVix [38]

Answer:

y = 2x

Step-by-step explanation:

Given that y varies directly as x then the equation relating them is

y = kx ← k is the constant of variation

To find k use the condition y = 0.8 when x = 0.4

k = $\frac{y}{x}$ = $\frac{0.8}{0.4}$ = 2

y = 2x ← equation of variation

7 0

4 years ago

Is the function continuous at x = -17

Pavlova-9 [17]

For a function to be continuous at an x-value, say -17, you need to make sure two things line up:

The limit from the left equals the limit from the right.

$\lim_{x \to -17^{-}} f(x) = \lim_{x \to -17^{+}} f(x)$

This limit equals the functions value.

$\lim_{x \to -17} f(x) = f(-17)$

The left hand limit involves the first piece, f(x) = 20x + 1:

$\begin{aligned} \lim_{x \to -17^{-}} f(x) &= \lim_{x \to -17^{-}} (20x+1)\\[0.5em]&= 20(-17)+1\\[0.5em]&= -339\endaligned}$

The right hand limit invovles the second piece, f(x) = -10x^2:

$\begin{aligned} \lim_{x \to -17^{+}} f(x) &= \lim_{x \to -17^{+}} (-10x^2)\\[0.5em]&= -10\cdot (-17)^2\\[0.5em]&= -2890\endaligned}$

Since the two one-sided limits don't match, the function is not continuous at x=-17.

3 0

3 years ago

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

What is another way to write this number 300+70+1/5+8/100

barxatty [35]

Answer:

370.28

Step-by-step explanation:

you add the two whole numbers then you add the two fractions and make them a decimal after that put it all together and u get the answer

3 0

3 years ago