Erica can run 1/6 of a kilometer her school is 3/4 of a kilometer away from her home at this this speed how long would it take erica to run home from school
Answers
Answer:
4 minutes 30 seconds
Step-by-step explanation:
Given:
Erica can run 1/6 of a kilometer in one minute.
Her school is 3/4 of a kilometer away from her home.
Question asked:
How long would it take Erica to run home from school = ?
Solution:
As here distance is given and also speed is given, we will find time taken to run from home to school by using,
Therefore, it takes 4 minutes 30 seconds Erica to run home from school.
Related Questions
How do I solve this?
A tile factory earns money by a flat fee for delivery and sales price per tile sold. For one customer the tile factory shipped 10,000 units of retail stores at the rate of $0.25 Per tile. The total tile factory earned from that sale was 3000 which of the following equation describes the revenue of the tile from this sale in terms of tiles Sold? A) y=0.25x+10,000B) y=0.25x+3,000C) y=0.25x+500D) y=0.25x+0.25
Help please i need the answer
Which equation represents the line that passes through (-8,11) andand (4, 7/2)?A) y = -5/8x + 6 B) y = -5/8x + 16 C) y = 15/2x - 49 D) y = -15/2x + 71
Answers
Answer:
las fracciones decimales son:
21/10 = 2,1
439/100 = 4,39
34/10 = 3,4
35/1000 = 0,035
851 / 1000 = 0,851
Step-by-step explanation:
Toda fraccion cuyo denominador (el numero de abajo) es multiplo de 10 es considerada una fraccion decimal. No importa si el numerador es mayor al denominador. Justamente el sistema decimal fue concebido para facilitar la multiplicacion y division de los numeros por algun multiplo de 10.
Answers
Answer:
length: 6 inches
width: 7 inches
Step-by-step explanation:
Let n represent the number. Then the length is (n+2) and the width is (2n-1). The product of these dimensions is the area:
42 = (n+2)(2n -1) = 2n^2 +3n -2
2n^2 +3n -44 = 0 . . . . subtract 42
(n -4)(2n +11) = 0 . . . . . factor
n = 4 . . . . . . . . we aren't interested in the negative solution
Length = 4+2 = 6 . . . inches
Width = 2·4 -1 = 7 . . . inches
Final answer:
The dimensions of the rectangle given that its area is 42 square inches, the length is 2 inches more than a number, and the width is 1 inch less than twice the same number, are 6 inches by 7 inches.
Explanation:
To solve this problem, let's define the unknown number as
x. According to the problem, the length of the rectangle is 2 inches more than x (so it's x + 2), and the width is 1 inch less than twice the number x (which makes it 2x - 1).
Now, we'll use the formula for the area of a rectangle, which is length times width: (x + 2) * (2x - 1) = 42.
Solve this equation by expanding the parentheses (2x^2 + 4x - x - 2 = 42), simplifying (2x^2 + 3x - 2 - 42 = 0), and rearranging (2x^2 + 3x - 44 = 0).
Using the quadratic formula, we find that the possible values of x are 4 and -5.5. However, a negative size doesn't make sense in this context, so x = 4 inches. That makes the length = 4 + 2 = 6 inches, and the width = 2*4 - 1 = 7 inches. Therefore, "the dimensions of the rectangle are 6 inches by 7 inches".
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Answers
67%=67% of 360
percent means parts out of 100 so
67%=67/100=0.67
'of' means multiply
67% of 360=0.67 times 360=241.2
round
241
answer is 241 degrees
Answers
Answer:
The answer is 2000
Step-by-step explanation:
Answers
x is 22 so the larger number 'x+6' is equal to :
There are 28 students in the larger class.
Let's assume the number of students in one class is x.
Since the other class has 6 more students, the number of students in the other class is x + 6.
The total number of students in both classes is 50.
Therefore, we can write the equation:
x + (x + 6) = 50
2x + 6 = 50
Subtracting 6 from both sides:
2x = 44
Dividing both sides by 2:
x = 22
So, the number of students in the larger class, which is x + 6, is:
22 + 6 = 28
Therefore, there are 28 students in the larger class.
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(a) Find the approximate length of the plank. Round to the nearest tenth of a foot.
(b) Find the height above the ground where the plank touches the wall. Round to the nearest tenth of a foot.
2. A support wire is strung from a tree. The bottom of the wire is 24 ft from the tree. The length of the wire is 30 ft. Find each of the following values, and show all your work.
(a) Find the value of x. Round to the nearest tenth of a degree.
(b) Find the approximate height where the wire touches the tree. Round to the nearest foot.
Answers
(a) Using sine rule
ground/Sin 49 = plank/Sin 90
But sine 90 = 1, ground = 3 ft
Then,
3/Sin 49 = plank
Length of the plank = 3/Sin 49 ≈ 4.0 ft (rounded to nearest tenth)
(b) Height where the plank touches the wall
wall = Sqrt (plank^2 - ground^2) = Sqrt (4.0^2 - 3.0^2) ≈ 2.6 ft (Rounded to nearest tenth)
Question 2:
(a) Angle x
ground = 24 ft
support wire = 30 ft
Applying sine rule
support wire/Sin 90 = ground/Sin (90-x) ----- but Sin 90 = 1
Then,
Support wire = ground/Sin (90-x)
Sin (90-x) = ground/support wire = 24/30 = 0.8
90-x = Sin^-1(0.8) = 53.13 => x = 90-53.13 = 36.9°
(b) Height where the wire touches the tree (tree)
tree = Sqrt (support wire^2 - ground^2) = Sqrt (30^2 - 24^2) = 18 ft