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viva [34]
3 years ago
14

HELP!!!

Mathematics
2 answers:
cestrela7 [59]3 years ago
8 0
Yeah, it’s all about calculating probabilities here. The probability on getting a 1 or 2 on a cube is 1/3 (2 options out of 6 total), whereas the other two options are either 50% or greater.
natali 33 [55]3 years ago
5 0

Answer: I would say rolling a number cube

Step-by-step explanation:

when flipping a coin you will have a 1/2 chance of landing on heads and then when spinning the spinner you would have a 3/4 chance since there is 3 numbers greater then 1 and then with the cube since it has 6 sides and your only getting 1 or two you would have a 2/6 chance.

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2. Find the product (2x + 3)(3x - 2).
eimsori [14]

Answer:

6x² + 5x - 6

Step-by-step explanation:

Given

(2x + 3)(3x - 2)

Each term in the second factor is multiplied by each term in the first factor, that is

2x(3x - 2) + 3(3x - 2) ← distribute both parenthesis

= 6x² - 4x + 9x - 6 ← collect like terms

= 6x² + 5x - 6

6 0
3 years ago
Read 2 more answers
Don't really understand how to solve please help <3​
valina [46]
This is how the first one would be done. sorry if it’s a little messy the answer is 9. but try using photomath for the rest. it gives you the answer and explains how to do it

8 0
3 years ago
The actual income for this month has been reduced by $200. How can this budget be modified so there will be a positive actual ne
alexira [117]
The positive actual net income is $100
6 0
3 years ago
Suppose the length of a side of<br> a square is s. Write a formula to find the<br> area of a square.
densk [106]

Answer:

A=sw

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
What numbers are zeros of g(x) = x^2 - 2x - 4? take your time if you need to!
ipn [44]

Answer:

x = 1 + \sqrt5, x = 1 - \sqrt5

Step-by-step explanation:

Hello!

We can solve the quadratic by using the quadratic formula.

Standard form of a quadratic: ax^2 + bx + c = 0

Quadratic Formula: x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}

Given our Equation: g(x) = x^2 - 2x - 4

  • a = 1
  • b = -2
  • c = -4

Plug the values into the equation and solve.

<h3>Solve</h3>
  • x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}
  • x = \frac{-(-2)\pm\sqrt{(-2)^2 - 4(1)(-4)}}{2(1)}
  • x = \frac{2\pm\sqrt{4 +16}}{2}
  • x = \frac{2\pm\sqrt{20}}{2}
  • x = \frac{2\pm\sqrt{4 * 5}}{2}
  • x = \frac{2\pm(\sqrt4 * \sqrt5)}{2}
  • x = \frac{2\pm2\sqrt5}{2}
  • x = 1 + \sqrt5, x = 1 - \sqrt5
7 0
2 years ago
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