## yeah_i_suck_at_math 9 months ago If (x + 5)^2 = 35, what is the approximate positive value of x? 0.92 2.65 3.16 5.48

1. yeah_i_suck_at_math

2. BillBell

$x + 5 = +/- \sqrt{35}$ Question asks for the positve value of x. So we have $x = -5 + \sqrt{35}$ because $x = -5 - \sqrt{35}$ is negative. You can use a calculator or google to get the approximate value of this.

3. kym02

I got A too but i'm not sure

4. BillBell

That's what I get too.

5. yeah_i_suck_at_math

yeah its A, thanks guys, can you help me on another question, its harder than this one

6. BillBell

7. kym02

sure

8. yeah_i_suck_at_math

Debbie sells cookies and hotdogs. The daily cost of making cookies is $460 more than the difference between the square of the number of cookies sold and 20 times the number of cookies sold. The daily cost of making hotdogs is modeled by the following equation: C(x) = 2x2 - 80x + 1,200 C(x) is the cost in dollars of selling x hotdogs. Which statement best compares the minimum daily cost of making cookies and hotdogs? It is greater for hotdogs than cookies because the approximate minimum cost is$300 for cookies and $392 for hotdogs. It is greater for hotdogs than cookies because the approximate minimum cost is$360 for cookies and $400 for hotdogs. It is greater for cookies than hotdogs because the approximate minimum cost is$400 for cookies and $360 for hotdogs. It is greater for cookies than hotdogs because the approximate minimum cost is$392 for cookies and \$300 for hotdogs.

9. yeah_i_suck_at_math

10. yeah_i_suck_at_math

or B, not sure though

11. BillBell

dC(x)/dx = 2x - 20 = 0 => x = 10 => minimum cost of cookies = 460 + 10^2 - 20(10) = 360 dH(x)/dx = 4x - 80 = 0 => x = 20 => minimum cost of hotdogs = 1200 + 2(20^2) - 80(20) = 400

12. yeah_i_suck_at_math

so it was B

13. BillBell

So it appears.

14. yeah_i_suck_at_math

ok and last one .. A quadratic equation is shown below: x2 - 8x + 13 = 0 Which of the following is the first correct step to write the above equation in the form (x - p)2 = q, where p and q are integers? Subtract 5 from both sides of the equation Subtract 3 from both sides of the equation Add 5 to both sides of the equation <-- my answer Add 3 to both sides of the equation

15. BillBell

Back in a few minutes ...

16. BillBell

$( x - p)^{2} = x^{2} - 2p + p ^{2}$ Compare the equation you have been given with this one. Then -2p = -8 and therefore p =4. With this value of p we get $x^{2} - 8x + 16$ and we want to make the original equation into this so we subtract 3 from both sides.

17. yeah_i_suck_at_math

yeah i already finished this question but thanks

18. yeah_i_suck_at_math

will you help me on a new question please ? @BillBell