Find 2x − 3y − z if x = −2, y = 3, and z = −2 −11 11 −15 15

I need help finding the domain and range of this graph

SIZIF [17.4K]

Answer:

Step-by-step explanation:

Okay so to find the domain and range we need to find the x values and the y values: the domain is x and the range is y. Therefore the start of our x values can be concluded as -5 and the end is 5, so our domain will be -5 < x < 5: Note, I couldn't put the less than or equal to symbol. SO now our range is the values that are up and down, the range does down at 1 and the range is the highest at 3 so we can determine the range is 1 < y < 3. This is a continuous graph, so there is continuous x values and y values in this graph, so it's simple to use the start and the end of the domain and range followed by the possible y and x values which is represented by y and x.

Hope this helps. :)

8 0

3 years ago

What is 938 written in standard form?

Furkat [3]

Answer:

$9.38 \times 10^{2}$

Step-by-step explanation:

All numbers in scientific notation or standard form are written

$m \times 10^n$ ,

where m is a number between 1 and 10.

8 0

3 years ago

Read 2 more answers

A1 Plumbing Service charges $35 per hour plus a $25 travel charge for a service call. Good Guys Plumbing Repair charges $40 per

AlladinOne [14]

A1 = 35x+25

Good guys = 40x

35x+25 = 40x

subtract 3x x from each side

25 = 5x

divide each side by 5

x = 25/5 = 5

so 5 hours they would be the same cost, anything over 5 hours and A1 would be better

so X>5 for A1 to be the better price

3 0

3 years ago

If a certain cannon is fired from a height of 8.8 meters above the ground, at a certain angle, the height of the cannonball ab

Dennis_Churaev [7]

Answer:

It would take approximately 6.50 second for the cannonball to strike the ground.

Step-by-step explanation:

Consider the provided function.

$h(t)=-4.9t^2+30.5t+8.8$

We need to find the time takes for the cannonball to strike the ground.

Substitute h(t) = 0 in above function.

$-4.9t^2+30.5t+8.8=0$

Multiply both sides by 10.

$-49t^2+305t+88=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are: $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute a = -49, b = 305 and c=88

$t=\frac{-305+\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}=-\frac{-305+\sqrt{110273}}{98}\\t = \frac{-305-\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}= \frac{305+\sqrt{110273}}{98}$