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3 years ago

Which equation could be used to solve this problem?

Mathematics

1 answer:

vredina [299]3 years ago

6 0

The equation that represents how many items are on each list is option B) x + (x-9) = 33.

Step-by-step explanation:

There are two different lists of grocery.

The number of items in the 1st grocery list = x

The number of items in the 2nd grocery list differs by 9 from the number of items in the 1st grocery list.

Therefore, the number of items in the 2nd grocery list = (x-9)

The total number of items for both lists = 33.

The equation can be formed as :

number of items in the 1st list + number of items in the 2nd list = total number of items.

⇒ x + (x-9) = 33

The equation that represents how many items are on each list is x + (x-9) = 33.

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On which number line are -3 and its opposite shown?

netineya [11]

Answer:

it would be 3

Step-by-step explanation:

your welcome

7 0

3 years ago

The annual tuition at a specific college was $20,500 in 2000, and $45,4120

nika2105 [10]

Answer: the tuition in 2020 is $502300

Step-by-step explanation:

The annual tuition at a specific college was $20,500 in 2000, and $45,4120 in 2018. Let us assume that the rate of increase is linear. Therefore, the fees in increasing in an arithmetic progression.

The formula for determining the nth term of an arithmetic sequence is expressed as

Tn = a + (n - 1)d

Where

a represents the first term of the sequence.

d represents the common difference.

n represents the number of terms in the sequence.

From the information given,

a = $20500

The fee in 2018 is the 19th term of the sequence. Therefore,

T19 = $45,4120

n = 19

Therefore,

454120 = 20500 + (19 - 1) d

454120 - 20500 = 19d

18d = 433620

d = 24090

Therefore, an

equation that can be used to find the tuition y for x years after 2000 is

y = 20500 + 24090(x - 1)

Therefore, at 2020,

n = 21

y = 20500 + 24090(21 - 1)

y = 20500 + 481800

y = $502300

6 0

3 years ago

Simplify u^2+3u/u^2-9 A.u/u-3, =/ -3, and u=/3 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

Use a graphing utility to graph the function and the damping factor of the function in the same viewing window.

Bas_tet [7]

Answer:

The function touches the damping factor

at x= $\frac{(4n-3)\pi}{2}$ and x= $\frac{(4n-1)\pi}{2}$