All categories

3 years ago

Describe how the graph of y | x| -4 is like the graph of y= |x| and how it is different.

Mathematics

2 answers:

kondor19780726 [428]3 years ago

5 0

Answer:

The graph of y=|x|-4 is the same as y=|x|.

Step-by-step explanation:

zhannawk [14.2K]3 years ago

4 0

The graph y=|x|-4 is obtained from the graph y=|x| dy moving down 4 units the graph y=|x| along the y-axis (see, if x=0, then for y=|x|, y=0 and for y=|x|-4, y=-4).

These two graphs have the same form.

You might be interested in

Determine the solution to f(x) = g(x) using the following system of equations: (5 points) f(x) = 3x − 23 g(x) = −4.5x + 7

jeka94

Answer:

(The solution is (4, -11).

Step-by-stp explanation:

Let f(x) = g(x) = y:

y = 3x − 23

y = -4.5x + 7 Subtract the second equation from the first to eliminate y:

0 = 7.5x - 30

7.5x = 30

x = 4

Plug this into the first equation:

y = 3(4) - 23

y = -11.

5 0

3 years ago

Please help. Felipe is planning a party. He has $50 to spend on sandwiches That cost $2.50 Per person , $25 to spend on drinks t

MArishka [77]

Answer:

347

Step-by-step explanation:

i added them all

8 0

3 years ago

A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0

4 years ago

Pythagorean theorem practice test

xeze [42]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Pls help me i need to do all my math for to day and i dont know how to do it -11x(x+12)

Vesna [10]

Answer:

If you want to simplify, -11 $x^{2}$ -132x

Step-by-step explanation:

4 0

3 years ago