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iragen [17]
3 years ago
9

Describe how you could draw a diagram for a problem finding the total length for two strings, 15 inches long and 7 inches long

Mathematics
1 answer:
Tema [17]3 years ago
6 0
Pay attention in class
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Which transformation will always map a parallelogram onto itself?
eimsori [14]

Answer:

C..

Step-by-step explanation:

A rotation of 180 degrees about the center of the parallelogram will  do this.

4 0
3 years ago
In triangle ABC, AC=13, BC=84, and AB=85. Find the measure of angle C
r-ruslan [8.4K]

Answer:

90°

Step-by-step explanation:

Angle C = arccos((84²+13²-85²)/(2×84×13))

= arccos(0/2184)

= arccos(0)

= 90°

Answered by GAUTHMATH

3 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
Barbara wants to buy some glittery heart-shaped beads to make friendship bracelets. The beads she wants come in a small pack wit
AleksAgata [21]

Answer:

I feel u

Step-by-step explanation:u need help too

5 0
3 years ago
In the diagram below, line d is parallel to line c.
My name is Ann [436]

Answer:

85° = ∠x {corresponding angles}

∠x = ∠1 {vertically opp. angles}

And, ∠2 + 103° = 180° ( co-interior angles)

∠2 = 180° - 103°

∠2 = 77°

Now, m∠1 + m∠2 = 85° + 77°

= 162°

Hemce, option A. is the right answer.

6 0
2 years ago
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