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3 years ago

8

Find the perimeter of the flag in the scale drawling. PLZ HELP I AM CONFUSED

1 answer:

chubhunter [2.5K]3 years ago

7 0

I don't see a photo attached

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Marissa is standing next to an 18-foot tall flagpole that is casting a 12 foot long shadow. If Masissa's shadow is 3 1/2 feet lo

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What are the roots of f(x) = x2 – 48? –48 and 48 –24 and 24 Negative 8 StartRoot 3 EndRoot and 8 StartRoot 3 EndRoot Negative 4

Ivanshal [37]

Answer:

$x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Step-by-step explanation:

$x=4\sqrt{3},\:x=-4\sqrt{3}$

Considering the expression

$x^2\:-\:48$

Solving

$x^2-48=0$

$x^2-48+48=0+48$

$x^2=48$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{48},\:x=-\sqrt{48}$

Solving

$x=\sqrt{48}$

= $\sqrt{2^4\cdot \:3}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$

= $\sqrt{3}\sqrt{2^4}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^m}=a^{\frac{m}{n}}$

$\sqrt{2^4}=2^{\frac{4}{2}}=2^2$

$=2^2\sqrt{3}$

$=4\sqrt{3}$

So,

$x=4\sqrt{3}$

Similarly,

$x=-\sqrt{48}=-4\sqrt{3}$

Therefore, $x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Keywords: roots, expression

Learn more about roots from brainly.com/question/3731376

#learnwithBrainly

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Point A is at (3, -8) and point B is at (7, 3). What is the midpoint of line segment AB?

allochka39001 [22]

Answer:

$(5, -5/2)$

Step-by-step explanation:

The midpoint of any line segment is the average of the x- and y-values of the two endpoints.

Therefore, the midpoint of $AB$ is $((3+7)/2, (-8+3)/2)=(10/2, -5/2)=(5, -5/2)$ .

So, the answer is $\boxed{(5, -5/2)}$ and we're done!

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