Answer:
<h2>QT = 21</h2><h2>SV = 41</h2>
Step-by-step explanation:
From the diagram, it can be seen that TR is parallel to RV. This means that TR = RV. Given TR = 17 and RV = 3x+2
3x+2 = 17
3x = 17-2
3x = 15
x = 5
RV = 3(5)+2 = 17
QV = 4x+1 = 4(5)+1
QV = 21
Using Pythagoras theorem on ΔQRV to get RQ
QV² = QR²+RV²
21² = QR²+17²
QR² = 21²-17²
QR = 12.33
Using Pythagoras theorem on ΔQRT to get QT
From ΔQRT,
QT² = QR²+TR²
QT² = 12.33²+17²
QR² = 152.0289+289
QT² = 441.0289
QT =21
Since TS = 9(5)-4 = 41
Using Pythagoras theorem on ΔTRS
From ΔTRS,
TS² = RS²+TR²
41² = RS²+17²
RS² = 41²-17²
RS² = 1392
RS = 37.31
Similarly Using Pythagoras theorem on ΔRSV
From ΔRSV,
SV² = RV²+RS²
SV² = 17²+37.31²
SV² = 1681.0361
SV = 41
Part 1)
we have
the scale is 
the distance on a map is 
we know that
The scale is equal to the distance on a map divided by the real distance
Let
x------> distance on a map
y-------> real distance
S-------> scale

In this part we have

Find the value of y

Substitute the values


Convert to kilometers

Part 2)
In this part we have

Find the value of x

Substitute the values


Convert to centimeters

therefore
the answer is
The distance is 
Answer:
The weight of an apple is 150g rounded to the nearest 5g
what is the lower bound weight of the apple
what is the upper bound of the weight of the apple
Step-by-step explanation:
The weight of an apple is 150g rounded to the nearest 5g
what is the lower bound weight of the apple
what is the upper bound of the weight of the apple