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Oksi-84 [34.3K]
2 years ago
11

On Saturday, Craig rode his bike 5/8 of a mile. On Saturday, he rode his bike 1/2 of a mile. Craig added 5/8+1/2=6/10and plotted

6/10 on his number line.

Mathematics
1 answer:
VashaNatasha [74]2 years ago
3 0
* Craig's answer is not reasonable because to add fractions the denominators must be the same.

** Total distance = 5/8 + 1/2 = 5/8 + 4/8 = 9/8 miles

*** Using the line number to prove the answer:
The line number that represents the problem is in the attached figure.
while the distance between 0 and 1 divided to 8 sections
to represent (5/8) count 5 sections from zero ⇒⇒⇒ point (a)
and to represent (1/2) it is the midpoint between 0 and 1 which mean it is 4 sections but it will be counted from point (a) so, adding 4 sections to point (a) the result will be the point (b)
So, counting from 0 to point (b) will give us 9 sections
and while one section represents (1/8)
So the total distance will be 9 * (1/8) = 9/8 which is agree with the result obtained before

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You have a square flower garden that has an area of 400 square yards. What is the length of one side of the garden?
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20 yards

Step-by-step explanation:

The area of a square is calculated by multiplying its height by its base. Because they are the same in a square, the area of a square is obtained by squaring one of its sides. Let's call the sides 'x'. Then we can express the area of the garden as x^2=400.

From here we simply need to square root both sides which gives us:

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Therefore the length of one side of the garden is 20 yards.

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An artist is creating a sculpture using bendable metal rods of equal length. One rod is formed into the shape of a square and an
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6 0
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Read 2 more answers
Let C be the closed, piecewise smooth curve formed by traveling in straight lines between the points (-2,1), (-2,-3), (1,-1) , (
Greeley [361]

The given points are the vertices of the quadrilateral

Q=\left\{(x,y)\mid-2\le x\le1,\dfrac{2x-5}3\le y\le\dfrac{4x+11}3\right\}

By Green's theorem, the line integral is

\displaystyle\int_C2xy\,\mathrm dx+xy^2\,\mathrm dy=\iint_Q\frac{\partial(xy^2)}{\partial x}-\frac{\partial(2xy)}{\partial y}\,\mathrm dA

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6 0
2 years ago
among a group of students 50 played cricket 50 played hockey and 40 played volleyball. 15 played both cricket and hockey 20 play
kondaur [170]

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

3 0
2 years ago
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