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2 years ago

Which is a solution of the inequality? p 4 �� 2(p �� 10) > 0

Mathematics

1 answer:

Mademuasel [1]2 years ago

6 0

I think b 26 is correct but dont quote me on that

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3 years ago

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

damaskus [11]

<u>ANSWER: </u>

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

<u>SOLUTION:</u>

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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2 years ago

Please, I really need help with this! What is the volume of the cone?

MariettaO [177]

V=1/3*πr² *h,
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3 years ago

The answers to the proof

Karolina [17]

Great work! But please show more of the picture

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2 years ago

12 + 4x > 24 or 10 - 12x > 22

Wewaii [24]

Answer:

10 - 12 × > 22 Yan Yong sagot

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2 years ago

Which is a solution of the inequality? p 4 ��� 2(p ��� 10) > 0

Which is a solution of the inequality? p 4 �� 2(p �� 10) > 0