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3 years ago

Using a deck of 52 playing cards, what are the odds against drawing a 6,7,8, or 9?

Mathematics

1 answer:

STatiana [176]3 years ago

6 0

first option 4/13

Keeping in mind a deck has 4 each of these cards...

4×4=16

16/52

4/13

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Identify the x-intercepts of the function below f(x)=x^2+12x+24

damaskus [11]

ANSWER:

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

SOLUTION:

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$

Hence the x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

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lidiya [134]

Answer: 13/21

Step-by-step explanation:

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Ad libitum [116K]

Answer:

89 meters

Step-by-step explanation:

We know it's 9 + "the height of the triangle".

Fortunately, the height of the triangle (let's call it y) can be computed from:

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y = 80 (or y = -80 but it doesn't make sense in geometry).

so x = y + 9 = 89;

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Answer:

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Step-by-step explanation:

When two or three are gathered in His name

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Step-by-step explanation:

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