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adoni [48]
3 years ago
6

Simplify the expression (-502) square root 0

Mathematics
1 answer:
Firdavs [7]3 years ago
7 0

Answer:

-502\sqrt 0 = 0

Step-by-step explanation:

Given

-502\sqrt 0

Required

Simplify

-502\sqrt 0

----------------------------------------

\sqrt 0 = 0

----------------------------------------

So:

-502\sqrt 0 = -502 * 0

-502\sqrt 0 = 0

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A master electrician earns $62 per hour. His apprentice earns $40 per hour. The master electrician works 3 hours more than the a
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X=hours master worked
y=hours apprentice worked


62x+40y=492

if master worked 3hours more than apprntice
x=3+y

sub 3+y for x

62(3+y)+40y=492
expand
186+62y+49y=492
186+102y=492
minus 186 both sides
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y=3

sub back
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Answers;

Question 1 answer: The first and last option.

Question 1 explanation: 2(4x + 2) is 4x = 12 = 2 = 14 x 2 = 28 and 8x + 4 is 8 x 4 = 32 + 4 = 36 which is 8 more than 28 then, 3x = 9 + 2 + 3 x 2 = 28.

2(4x + 2) = 28 and 8x + 4 equals 36 which is 8 more and they're both equivalent to 2(3x + 2 + x) because 2(3x = 2 = x) equals 28.

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2 years ago
A patient takes vitamin pills. Each day he must have at least 420 IU of vitamin A, 5 mg of vitamin
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Answer:

minimum cost is 70 cents when Greg buys 3 pill type 1, and 2 pills type 2.

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I don't know if this is right... please someone help mee
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For the first circle, let's use the pythagorean theorem

\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{8^2+15^2}\implies c=\sqrt{289}\implies c=17

now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.

 so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.

when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.



now, let's check for second circle

\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{11^2+14^2}\implies c=\sqrt{317}\implies c\approx 17.8044938

well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.



let's check the third circle

\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{33^2+56^2}\implies c=\sqrt{4225}\implies c=\stackrel{33+32}{65}

this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.
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