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jok3333 [9.3K]
3 years ago
8

Please help I don’t really understand how to go about it.

Mathematics
1 answer:
Airida [17]3 years ago
3 0
\bf \textit{sum of an arithmetic sequence}
\\\\
S_n=\cfrac{n(a_1+a_n)}{2}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
----------\\
S_n=795\\
a_1=102\\
a_n=57
\end{cases}
\\\\\\
795=\cfrac{n(102+57)}{2}\implies 1590=159n
\\\\\\
\cfrac{1590}{159}=n\implies 10=n\\\\
-------------------------------

so the nth term is really the 10th term, and we know that's 57, thus

\bf n^{th}\textit{ term of an arithmetic sequence}
\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=10\\
a_{10}=57\\
a_1=102
\end{cases}
\\\\\\
57=102+(10-1)d\implies 57=102+9d\implies -45=9d
\\\\\\
\cfrac{-45}{9}=d\implies -5=d

so, that's the common difference... .so you'd surely know what the 3rd term is, notice the first one is 102.
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Which statement proves that △XYZ is an isosceles right triangle?
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To determine whether two lines are perpendicular, their slopes must be negative reciprocals of each other.

Solving for the slopes of XZ and XY:
slope of XZ = (y₂-y₁) / (x₂-x₁) = (6-3)/(5-1) = 3/4
slope of XY = (y₂-y₁) / (x₂-x₁) = (3--1)/(1-4) = -4/3
-4/3 is the negative reciprocal of 3/4, therefore XZ and XY are perpendicular to each other.

Solving for distance:
XZ = √[(y₂-y₁)²+(x₂-x₁)²] = √[(6-3)²+(5-1)²] = 5
XY = √[(y₂-y₁)²+(x₂-x₁)²] = √[(3--1)²+(1-4)²] = 5

<span>The slope of XZ is 3/4, the slope of XY is -4/3, and XZ = XY = 5. </span>
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A lioness hunting antelope runs a distance of 60 m in 15 s. Then she stops to survey the herd for 20 s. Next, she trots for 25 s
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Answer: The average speed is 3m/s.

Step-by-step explanation:

We can define the average speed as the quotient between the total distance traveled in a given period of time, and that period of time,

Here we have that:

d = distance

t = time.

"A lioness hunting antelope runs a distance of 60 m in 15 s"

1) d = 60m, t = 15s

"Then she stops to survey the herd for 20 s"

We add 0 meters to the distance, and 20 seconds to the time.

2) d = 60m + 0m, t = 15s + 20s = 35s

"she trots for 25 s over 60 m around the edge of the herd."

we add 60m to the distance, and 25s to the time

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"Finally, the lioness races after the antelope, traveling 120 m in 20 s"

we add 120m to the distance and 20s to the time.

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Then the average speed of the lioness in that period of time is:

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