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Pani-rosa [81]
3 years ago
7

Evaluate: 54-75+81-(-27)+53

Mathematics
2 answers:
romanna [79]3 years ago
4 0

Hi the answer is 344

solmaris [256]3 years ago
4 0

Answer:

140

Step-by-step explanation:

54-75=-21

-21+81=60

60-(-27)=87 or 60+27=87

87+53=140

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Kruka [31]

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3 0
3 years ago
We are two integers whose product is -36 and whose sum is -16 what are we?
SCORPION-xisa [38]
-18 and +2
;;;;;;;;;;;;;;;;;;
6 0
3 years ago
Suppose you are walking home after school. The distance from your home to school is 5 km. on foot, you can get home in 1/2 mins.
Sholpan [36]

Answer:

speed is distance over time.so you convert 1/2 to hours then solve

Step-by-step explanation:

hope it helped

7 0
3 years ago
I don't know if this is right... please someone help mee
worty [1.4K]
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.
6 0
3 years ago
I need help with this problem. thank you so much who ever answer this
skelet666 [1.2K]

-36/8 = -4.5

 -4.5  = -4.5

7 0
3 years ago
Read 2 more answers
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