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schepotkina [342]

3 years ago

15

How many groups of Two-thirds are in Three-fifths?

2 answers:

zmey [24]3 years ago

6 0

Answer:

0.9 or 9/10

Step-by-step explanation:

If you set up the equation 2/3x = 3/5, and solve for x you will get 9/10. In this equation, x is the number of groups. So, 2/3 times the number of groups, or x, is equal to 3/5. If you solve for x you get 0.9. Another, less confusing way to solve these types of equations is by dividng 3/5 by 2/3. If you divide 3/5 by 2/3, you will get how many groups of 2/3 are in 3/5.

cestrela7 [59]3 years ago

3 0

Answer:

9/10 of a group

Step-by-step explanation:

Take 3/5 and divide by 2/3

3/5 ÷ 2/3

Copy dot flip

3/5 * 3/2

9/10

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Answer:

Part A) The length of a straight line between the school and the Fire Station is $4.6\ miles$

Part B) The length of a straight line between the school and the hospital is $2.1\ miles$

Step-by-step explanation:

<u><em>The complete question in the attached figure</em></u>

Let

The positive x-coordinates ----> East

The positive y-coordinates ----> North

The negative x-coordinates ----> West

The negative y-coordinates ----> South

take the point A (0,0) as the Middle School (reference point)

Part A) we have

Town Hall is located 4.3 miles directly east of the Middle School

so

The coordinates of Town Hall are B(4.3,0)

The Fire Station is located 1.7 miles directly North of Town Hall

so

The coordinates of Fire Station are C(4.3,1.7)

What is the length of a straight line between the school and the Fire Station?

Remember that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

A(0,0) and C(4.3,1.7)

substitute

$d=\sqrt{(1.7-0)^{2}+(4.3-0)^{2}}$

$d=\sqrt{(1.7)^{2}+(4.3)^{2}}$

$d=4.6\ miles$

Part B) we have that

The hospital is 3.1 miles west of the fire station

so

The coordinates of the hospital are D(4.3-3.1,1.7)

D(1.2,1.7)

What is the length of a straight line between the school and the hospital?

Remember that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

A(0,0) and D(1.2,1.7)

substitute

$d=\sqrt{(1.7-0)^{2}+(1.2-0)^{2}}$

$d=\sqrt{(1.7)^{2}+(1.2)^{2}}$

$d=2.1\ miles$

7 0

3 years ago