Answer: her speed on the bicycle is 8 mph
Step-by-step explanation:
Let x represent the speed at which she walks. If she rides 6 mph faster than she can walk, then the speed at which she rides is (x + 6) mph
Distance = speed × time
Eli walk from her house to a friends house in one hour. This means that the distance covered is 1 × x = x miles.
Distance on her bicycle in 15 minutes. Converting 15 minutes to hours, it becomes 15/60 = 0.25 hour. Distance covered while riding is 0.25(x + 6) = 0.25x + 1.5
Since the distance is the same,then
x = 0.25x + 1.5
x - 0.25x = 1.5
0.75x = 1.5
x = 1.5/0.75
x = 2
her speed on the bicycle is
x + 6 = 2 + 6 = 8 mph
(a)
421₅ = 4×5² + 2×5¹ + 1×5⁰
… = 100 + 10 + 1
… = 111
(b)
11101₂ = 1×2⁴ + 1×2³ + 1×2² + 0×2¹ + 1×2⁰
… = 16 + 8 + 4 + 0 + 1
… = 29
(c)
13T₁₂ = 1×12² + 3×12¹ + 10×12⁰
… = 144 + 36 + 10
… = 190
Answer:
Step-by-step explanation:
first, you need to know the formulas for volumes for a sphere, cone, and cube.
sphere volume is defined: V=4/3πr^3
cone volume is defined: V=πr^2(h
/3)
cuboid volume is defined: V=lwh
Then just plug in:
sphere volume=4/3π6^3=904.78
cone volume=π5^2(13
/3)=340.34
cuboid volume=11*6*5=330
Answer:
perimeter is 4 sqrt(29) + 4pi cm
area is 40 + 8pi cm^2
Step-by-step explanation:
We have a semicircle and a triangle
First the semicircle with diameter 8
A = 1/2 pi r^2 for a semicircle
r = d/2 = 8/2 =4
A = 1/2 pi ( 4)^2
=1/2 pi *16
= 8pi
Now the triangle with base 8 and height 10
A = 1/2 bh
=1/2 8*10
= 40
Add the areas together
A = 40 + 8pi cm^2
Now the perimeter
We have 1/2 of the circumference
1/2 C =1/2 pi *d
= 1/2 pi 8
= 4pi
Now we need to find the length of the hypotenuse of the right triangles
using the pythagorean theorem
a^2+b^2 = c^2
The base is 4 ( 1/2 of the diameter) and the height is 10
4^2 + 10 ^2 = c^2
16 + 100 = c^2
116 = c^2
sqrt(116) = c
2 sqrt(29) = c
Each hypotenuse is the same so we have
hypotenuse + hypotenuse + 1/2 circumference
2 sqrt(29) + 2 sqrt(29) + 4 pi
4 sqrt(29) + 4pi cm
Answer:
Step-by-step explanation:
To polygons are said to be similar if all the corresponding angles of given polygons are equal.
In the given figure, it can be seen that in quadrilateral FSHB and quadrilateral KTWJ all the corresponding angles are equal.
∠F=∠K
∠S=∠T
∠H=∠W
∠B=∠J
Therefore, The given polygons are similar.
Hence,