Hey there!
<em>C = 0.25x + 40</em>
(a) For this part we solve for C when x = 240. Substitute 240 in for x.
C = 0.25(240) + 40
Multiply 240 by 0.25.
C = 60 + 40
Add.
C = 100
The cost is $100.
<u>Check</u>
100 = 0.25(240) + 40
100 = 60 + 40
100 = 100
(b) For this part we solve for x when C = 180. Substitute 180 for C.
180 = 0.25x + 40
Subtract 40 from each side.
140 = 0.25x
Divide each side by 0.25.
560 = x
This person drove 560 miles.
<u>Check</u>
180 = 0.25(560) + 40
180 = 140 + 40
180 = 180
(c) For this we must make an inequality with C = 150.
Since they want the cost to be <em>no more than</em> 150, we will use the ≥ symbol.
C ≥ 0.25x + 40
Substitute 150 for C. Solve for x.
150 ≥ 0.25x + 40
Subtract 40 from each side.
110 ≥ 0.25x
Divide each side by 0.25.
440 ≥ x
The max number of miles they can drive is 440.
<u>Check</u>
150 ≥ 0.25(440) + 40
150 ≥ 110 + 40
150 ≥ 150
Hope this helps!
0.03x + 15 y = 0.03x - 15 y = -0.03x + 15 y = -0.03x - 15
Q. How many triangles can be constructed with sides measuring 5 m, 16 m, and 5 m?
Solution:
Here we are given with the sides of the triangle as 5m, 16m and 5.
As the Triangle inequality we know that
The sum of the length of the two sides should be greater than the length of the third side. But this inequality fails here.
Hence no triangle can be made.
So the correct option is None.
Q.How many triangles can be constructed with sides measuring 6 cm, 2 cm, and 7 cm?
Solution:
Here we are given with the sides of the triangle as 6m, 2m and 7m.
As the Triangle inequality we know that
The sum of the length of the two sides should be greater than the length of the third side. The given values follows the triangle inequality.
Hence one triangle can be formed.
So the correct option is one.
Answer:
8
Step-by-step explanation:
The radius is part of a 3-4-5 special right triangle.
The diameter is two times the radius, so 4*2=
<em>I hope this helps! :)</em>
Answer:
3.375 miles
Step-by-step explanation:
Each day Maria walks 0.75 miles, so if she walks for 4.5 days, then 0.75*4.5= 3.375 miles.
