Solve for t. 50= − 10 t + 80

Answers

Answer 1
Answer: The answer is t= 3 !!!
Answer 2
Answer: 50= - 10t+80
We move all terms to the left:
50 - ( -10t+80) =0
Get rid of the parentheses
10t - 80+50=0
We add all the numbers together, and all the variables
10+ - 30=0
We move all the terms containing t to the left, all the other terms to the right
10t = 30
t = 30/10
t = 3

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Multiply (2m + 3)(m2 – 2m + 1).

Answers

Are you simplifying?

The length of a rectangle is 5 cm more than twice the width the perimeter of the rectangle is 34 cm what are the dimensions

Answers

If the width is 4, we're told that the length is 5 more than twice the width, or (2*4)+5, which is 13.

How many minutes in a day

Answers

If an hour has 60 minutes and a day has 24 hours , there are 60 times 24 in a day

60*24 = 1440


There are 1440 minutes in a day !


I hope that's help !

What is -2+6y=18 3y=15 ?

Answers

-2+6y=18\ \ \  |add\ 2\ to\ both\ sides\n6y=20\ \ \ \ \ |divide\ both\ sides\ by\ 6\ny=(20)/(6)\n\n\boxed{y=(10)/(3)}\n---------------------\n\n3y=15\ \ \ \ \ |divide\ both\ sides\ by\ 3\n\n\boxed{y=5}
\left \{ {{-2+6y=18} \atop {3y=15}} \right.  \n +2+2 \left \{ {{6y=20} \atop {3y=15}} \right. \n /3/3 \left \{ {{6y=20} \atop {y=5}} \right.\n  /6/6 \left \{ {{y=3 (1)/(3) } \atop {y=5}} \right.

|2x-3| < 9 solve for x

Answers

|2x-3| < 9 \n \n2x-3 < 9 \ \ and \ \ 2x-3 > -9 \n \n 2x < 9 +3 \ \ and \ \ 2x > -9 +3 \n \n 2x < 12\ \ and \ \ 2x > -6 \n \n x < 6\ \ and \ \ x > - 3 \n \n x \in (-3,6)
|2x-3| < 9\iff2x-3 < 9\ \wedge\ 2x-3 > -9\n\n2x < 9+3\ \wedge\ 2x > -9+3\n\n2x < 12\ \wedge\ 2x > -6\n\nx < 6\ \wedge\ x > -3\n\n\nx\in(-3;\ 6)

Use the discriminant to determine the number and type of solutions to the quadratic equation. Show all work for full credit! 2+ 8= 13

Answers

Answer:

The equation has two different real solutions

Step-by-step explanation:

The discriminant of the quadratic equation ax² + bx + c = 0 is Δ = b² - 4ac, it used to find the number and type of solutions

  • Δ > 0, means two different real solutions
  • Δ = 0, means one real solution
  • Δ < 0, means no real solutions

∵ The equation is a² + 8a = 13

- Put it in the form ax² + bx + c = 0

- Subtract 13 from both sides

∴ a² + 8a - 13 = 0

∴ The coefficient of a² = 1, the coefficient of a = 8 and the

   numerical term = -13

∵ Δ = (coefficient of a)² - 4(coefficient of a²)(numerical term)

∴ Δ = (8)² - 4(1)(-13)

∴ Δ = 64 + 52

∴ Δ = 116

∵ Δ > 0

The equation has two different real solutions