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3 years ago

Subtract 25.45 from 51.82. Give your answer to 2 decimal places.

Mathematics

1 answer:

Usimov [2.4K]3 years ago

4 0

Answer:

Answer is, 26.4

Step-by-step explanation:

By subtracting 51.82 by 26.37, you're left with a total of 26.37, and by rounding that to the tenths place, you get 26.4

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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

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

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

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