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Masteriza [31]
3 years ago
6

*20 points* Complete the proof. Prove: △TUW ≅ △SVW

Mathematics
2 answers:
Sergeu [11.5K]3 years ago
8 0

Answer:

<h2>B) ASA theorem.</h2>

Step-by-step explanation:

The missing reason is ASA, which is about the congruence between Angle-Side-Angle, like this case.

Notice that we already have to pair of angles congruent, and one pair side congruent, that's why we use ASA to demonstrate the congruence.

Therefore, the right answer is B.

Ghella [55]3 years ago
6 0

B) ASA Congruence

Hope this helps!

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The student government association at a certain college assigns a number to each student enrolled, puts the numbers in order, an
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Describe the continuity of the graphed function
marusya05 [52]

we know that

Any function f(x) is continuous at x=a only if

\lim_{x \to a-} f(x) = \lim_{x \to a+} f(x)=f(a)

We can see that this curve is smooth everywhere except at x=3

so, we will check continuity at x=3

Left limit is:

\lim_{x \to 3-} f(x) = -\infty

Right limit is:

\lim_{x \to 3+} f(x) = +\infty

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f(3)= DNE

we can see that left limit is not equal to right limit

so, limit does not not exist

so, this function is discontinuous at x=3

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5 0
3 years ago
The following table shows the speeds at which Jeff does 4 activities. For example, biking has a rate of 19 miles in 2 hours. Whi
ExtremeBDS [4]

The type of activity that Jeff does at a speed of 4.5 miles per hour is Swimming

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem !

<em>This problem is about Kinematics.</em>

We will solve it in the following way

\texttt{Speed of Biking} = \texttt{Distance of Biking} \div \texttt{Time Taken}

\texttt{Speed of Biking} = \texttt{19} \div \texttt{2}

\texttt{Speed of Biking} = 9.5 \texttt{ miles per hour}

\texttt{ }

\texttt{Speed of Walking} = \texttt{Distance of Walking} \div \texttt{Time Taken}

\texttt{Speed of Walking} = \texttt{14} \div \texttt{4}

\texttt{Speed of Walking} = 3.5 \texttt{ miles per hour}

\texttt{ }

\texttt{Speed of Running} = \texttt{Distance of Running} \div \texttt{Time Taken}

\texttt{Speed of Running} = \texttt{15} \div \texttt{3}

\texttt{Speed of Running} = 5 \texttt{ miles per hour}

\texttt{ }

\texttt{Speed of Swimming} = \texttt{Distance of Swimming} \div \texttt{Time Taken}

\texttt{Speed of Swimming} = \texttt{18} \div \texttt{4}

\texttt{Speed of Swimming} = \boxed{4.5 \texttt{ miles per hour}}

\texttt{ }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

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