3 years ago

Need help on this math question

Mathematics

1 answer:

8090 [49]3 years ago

7 0

= 2^-16 * 3^10 * 3^-8 * 3^-12 * 2^28

= 2^(-16 + 28) * 3^(10 - 8 - 12)

= 2^12 * 3^-10

= 2^12 / 3^10

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Justify your reasoning when solving the equation: 3(x+4)=8x-18

vampirchik [111]

Answer:

Step-by-step explanation:

3x plus 12= 8x -18

subtract twelve on both sides:

3x= 8x- 30

minus 8x on both sides:

-5= -30

divide:6

3 0

2 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

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

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