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3 years ago

Can someone explain this problem? I don't understand it...

Mathematics

1 answer:

Irina18 [472]3 years ago

8 0

1) A function in when there is only one y for that x. (In other words it passes the vertical line test.)

2) Continuous means there are no breaks.

3) For x values less than -2, the graph is going up.

4) Between x values -2 and 8, the graph is going down.

5) After the x value 8, the graph starts to increase again.

6) The graph should only pass the x axis at 5, and only pass the y axis at 2.

Hope this helps.

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Need help solving x

viva [34]

The coefficient of x in the equation is 1/(300 grams). Multiply by the reciprocal of that, which is (300 grams).

x = (300 grams)×(2 grams)/(100 grams) = (300×2/100) grams = 6 grams

_____

This is the way you solve any equation where x is multiplied by some value in the only term on one side of the equal sign.

5 0

3 years ago

What are the vertex, axis of symmetry, domain and range of the function:

DiKsa [7]

Y=(-x*x)-4x+3
y=-(x^2+4x-3)
4+y=-(x^2+4x+4)+3
y+4=-(x-2)^2.) +3
-4. -4
y=-((x-2)^2)-1

Answers:
Vertex: (2,-1)
AOS:x = 2
Domain: All Real Numbers
Range:[-1,Infinity)

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3 years ago

Help me please<br> Idk what to do

Ber [7]

Answer:

x=36

Step-by-step explanation:

First rewrite equation then multiply each side by 36. 9x-36=4x+144.

then move the variable 9x-4x+36=144. Next subtract 9x and 4x ...5x=144+36.... 5x=180. Lastly you divide 180÷5 this your answer x=36

4 0

3 years ago

HELP!!! WILL GIVE BRAINLIEST!!!! The agriculture department of a university works with individuals wanting to build greenhouses

Zanzabum

Answer:

D ... y= 3,450/ 1=10.13e^-0.2854

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Simplify u^2+3u/u^2-9<br> A.u/u-3, =/ -3, and u=/3<br> B. u/u-3, u=/-3

VashaNatasha [74]

  The correct answer is: Answer choice: [A]:
__________________________________________________________
→  " $\frac{u}{u-3}$ " ; " { u $\neq$ ± 3 } " ;

  →  or, write as: " u / (u − 3) " ;  {" u ≠ 3 "}  AND: {" u ≠ -3 "} ;
__________________________________________________________
Explanation:
__________________________________________________________
We are asked to simplify:

   $\frac{(u^2+3u)}{(u^2-9)}$ ;

Note that the "numerator" —which is: "(u² + 3u)" — can be factored into:
  → " u(u + 3) " ;

And that the "denominator" —which is: "(u² − 9)" — can be factored into:
  → "(u − 3) (u + 3)" ;
___________________________________________________________
Let us rewrite as:
___________________________________________________________

→ $\frac{u(u+3)}{(u-3)(u+3)}$ ;

___________________________________________________________

→ We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ;  since:

" $\frac{(u+3)}{(u+3)} = 1$ " ;

→ And we have:
_________________________________________________________

→  " $\frac{u}{u-3}$ " ; that is: " u / (u − 3) " ; { u $\neq 3$ } .
and: { u $\neq-3$ } .

→ which is:  "Answer choice: [A] " .
_________________________________________________________

NOTE:  The "denominator" cannot equal "0" ; since one cannot "divide by "0" ;

and if the denominator is "(u − 3)" ; the denominator equals "0" when "u = -3" ; as such:

"u $\neq$ 3" ;

→ Note: To solve: "u + 3 = 0" ;

Subtract "3" from each side of the equation;

→ " u + 3 − 3 = 0 − 3 " ;

→ u = -3 (when the "denominator" equals "0") ;

→ As such: " u $\neq$ -3 " ;

Furthermore, consider the initial (unsimplified) given expression:

→   $\frac{(u^2+3u)}{(u^2-9)}$ ;

Note:  The denominator is: "(u²  − 9)" .

The "denominator" cannot be "0" ; because one cannot "divide" by "0" ;

As such, solve for values of "u" when the "denominator" equals "0" ; that is:
_______________________________________________________

→ " u² − 9 = 0 " ;

→ Add "9" to each side of the equation ;

→ u² − 9 + 9 = 0 + 9 ;

→ u² = 9 ;

Take the square root of each side of the equation;
to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ;

→ √(u²) = √9 ;

→ | u | = 3 ;

→ " u = 3" ; AND; "u = -3 " ;

We already have: "u = -3" (a value at which the "denominator equals "0") ;

We now have "u = 3" ; as a value at which the "denominator equals "0");

→ As such: " u $\neq 3$ " ; "u $\neq$ -3 " ;

or, write as: " { u $\neq$ ± 3 } " .

_________________________________________________________

6 0

3 years ago