Using only odd numbers for numerators, write two different subtraction problems that have a difference of 1/2.

You need to figure out how much money you make each week at the sales job. Assume you work a

Over [174]

Answer:

This true because why ?

Step-by-step explanation:

3 0

2 years ago

Please tell me if my answer is right <br> NO LIES brainlest<br> report you then

ladessa [460]

Answer:

Your answers are correct.

Step-by-step explanation:

The "<" sign means less than and the ">" sign means greater than. It is clear that you have a handle on them.

1.) If Steve has no more that 28 toys, then the less than sign would represent the fact that Steve can have a maximum of 28 toys, and no more.

2.) If the temperature is more that 28 degrees Fahrenheit, then the greater than sign is appropriate in order to imply that the variable represents a higher temperature.

3.) If tony is younger than 29 years old, then the less than sign fits to show that the variable represents an age that is smaller than 29.

4.) If the table is greater than 29 kilograms, then the greater than sign is needed. It shows that the table has a greater weight than 29 kilograms.

Hope this helps! :D

4 0

2 years ago

Is this function linear or nonlinear?

vichka [17]

Lineare means it is in a straight line aka highest exponent on any placeholder is 1

y=2^2-4
y^1=2^2-4
yep

answer is this is linear

5 0

3 years ago

How do I solve 2f+10?

Scorpion4ik [409]

First of all, you want the variable to be isolated, so in order to do that, you divide in both sides. Ten divided by two is five. So f = 5.

2f + 10
--- ---
2 2

10 divided by 2 = 5

5 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]:
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→  " $\frac{u}{u-3}$ " ; " { u $\neq$ ± 3 } " ;

  →  or, write as: " u / (u − 3) " ;  {" u ≠ 3 "}  AND: {" u ≠ -3 "} ;
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Explanation:
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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)" ;
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Let us rewrite as:
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→ $\frac{u(u+3)}{(u-3)(u+3)}$ ;

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

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

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6 0

3 years ago