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3 years ago

Please help me with this question (if you can)! Thank you so much, help is very much appreciated! :)

Mathematics

1 answer:

Rasek [7]3 years ago

7 0

The answer for your question is C because the first cup is decaf and second one is not.

Hope this helps:)

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Greg feeds his dog 2/3 pounds of dog food for each meal. Greg has just bought 18 pounds of dog food. For how many meals will the

yan [13]

We can divide 18 pounds by 2/3 pounds to find how many servings of meals the dogs will have.
We will use the keep change flip rule to solve this.
18 / 2/3 is then the same as 18 x 3/2
We then multiply across to get
54/2
This would be 27 meals that the food will last

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3 years ago

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3 years ago

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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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3 years ago