Answer:
a) 6 mins
b) 70km/h
c) t= 45
Step-by-step explanation:
a) The bus stops from t=10 to t=16 minutes since the distance the busvtravelled remained constant at 15km
Duration
= 16 -10
= 6 minutes
b) Average speed
= total distance ÷ total time
Total time
= 24min
= (24÷60) hr
= 0.4 h
Average speed
= 28 ÷0.4
= 70 km/h
c) Average speed= total distance/ total time
Average speed
= 80km/h
= (80÷60) km/min
= 1⅓ km/min
1⅓= 28 ÷(t -24)
<em>since</em><em> </em><em>duration</em><em> </em><em>for</em><em> </em><em>return</em><em> </em><em>journey</em><em> </em><em>is</em><em> </em><em>from</em><em> </em><em>t</em><em>=</em><em>2</em><em>4</em><em> </em><em>mins</em><em> </em><em>to</em><em> </em><em>t</em><em> </em><em>mins</em><em>.</em>
(t -24)= 28
t - 32= 28
t= 32 +28
t= 60
t= 
t= 45
*Here, I assume that this is a displacement- time graph, so the distance shown is the distance of the bus from the starting point because technically if it is a distance-time graph, the distance would still increase as the bus travels the 'return journey'.
Thus, distance is decreasing after t=24 and reaches zero at time= t mins so that is the return journey. (because when the bus returns back to starting point, displacement/ distance from starting point= 0km)
The length of UV is 11.
Solution:
UV = 2x - 13
VW = -18 + 2x
UW = 17
UV + VW = UW
2x - 13 - 18 + 2x = 17
4x - 31 = 17
Add 31 on both sides.
4x - 31 + 31 = 17 + 31
4x = 48
Divide by 4 on both sides.
x = 12
UV = 2x - 13
= 2(12) - 13
= 24 - 13
UV = 11
The length of UV is 11.
Answer:
11.2
Step-by-step explanation:
two triangles are similar
so, 14/25 = x/20
Answer:
Please check the explanation!
Step-by-step explanation:
Important factors to remember regarding the like terms:
- 'Like terms' are terms whose variables such as 2x and 6x are the same.
- Also if the variables contain the exponents such as 4 in x⁴, they must be the same too.
- Since the coefficient doesn't affect likeness, therefore we can say that all constant terms would be treated as the like terms.
Question 7
Given the terms
2b b⁶ b x⁴ 3b⁶ 2x²
From the list of terms, there are two groups of like terms which are:
2b b (Like terms containing the variable b)
b⁶ 3b⁶ (Like terms containing the variable b with the same exponents)
- The remaining two x⁴ and 2x² terms are different from each other and the rest of the terms
Question 8
Given the terms
6 2n 3n² 3m² n/4 7
From the list of terms, there is one group of like terms which are:
6 7 (Constants Like terms)
2n n/4 (The terms with the same variable n)
- All the remaining terms such as 3n², 3m² are all different from each other and the rest of the terms.
Question 9
Given the terms
10k² m 3³ p/6 2m
From the list of terms, there are there is one group of like terms which are:
m 2m (The terms withe same variable m)
- All the remaining terms such as 10k², 3³, p/6 are all different from each other and the rest of the terms.
Question 10
Given the terms
6³ y³ 3y² 6² y 5y³
From the list of terms, there are two groups of like terms which are:
y³ 5y³ (The terms withe same variable m and exponent)
6³ 6² (Constant like terms. Plz don't confuse the power as all are the same constant terms)
- All the remaining terms such as y and 3y² are all different from each other and the rest of the terms.
Answer medical
Step-by-step explanation:
