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3 years ago

If you spin a 5 section spinner and toss a coin, how many outcomes are possible?

Mathematics

1 answer:

baherus [9]3 years ago

7 0

Answer:

Step-by-step explanation:

They are independent events, so they don't influence each other. So it's just 5 possibilities on the spinner times 2 possibilities on the coin, 10.

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1. Solve for x: 3(2x – 1) – 10 = 8 + 5x A. -7 B. -3 C. 19 D. 21

user100 [1]

Answer:

(D) 21

Explanation:

3(2x−1)−10=8+5x

Step 1: Simplify both sides of the equation.

3(2x−1)−10=8+5x

(3)(2x)+(3)(−1)+−10=8+5x(Distribute)

6x+−3+−10=8+5x

(6x)+(−3+−10)=5x+8(Combine Like Terms)

6x+−13=5x+8

6x−13=5x+8

Step 2: Subtract 5x from both sides.

6x−13−5x=5x+8−5x

x−13=8

Step 3: Add 13 to both sides.

x−13+13=8+13

x=21

3 0

4 years ago

Read 2 more answers

Write each decimal as a percent.

Mumz [18]

1. 74%
2. 15%
3. 9%
4. 61.7%
5. 83.4%

Hope this helps

4 0

3 years ago

Read 2 more answers

Ben needs to mix 1.5 cups of lemonade concentrate with 4 cups or water to make lemonade. He has 6 cups of concentrate. How much

lubasha [3.4K]

Answer:

24

Step-by-step explanation:

To get your answer, you need to multiply 1.5 and 6.

1.5 x 6 = 9

Then you multiply 4 and 6.

4 x 6 = 24

Which is your final answer.

<h3>I hope this helps you! Good Luck!</h3>

4 0

3 years ago

PLEASE HELP!!!!!!!!!!!!! DUE SOON

Sergio039 [100]

$\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ h=-16t^2+\stackrel{\stackrel{v_o}{\downarrow }}{65}t$

now, take a look at the picture below, so for 2) and 3) is the vertex of this quadratic equation, 2) is the y-coordinate and 3) the x-coordinate.

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+65}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{65}{2(-16)}~~,~~0-\cfrac{65^2}{4(-16)} \right) \implies \left( \cfrac{65}{32}~,~0- \cfrac{4225}{-64}\right)$

$\bf \left( \cfrac{65}{32}~,~0+ \cfrac{4225}{64}\right)\implies \left( \stackrel{seconds}{2\frac{1}{32}}~~,~~ \stackrel{feet~hight}{66\frac{1}{64}}\right)$