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2 years ago

Part B

Mathematics

1 answer:

Troyanec [42]2 years ago

7 0

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:

The complete question in the attached figure

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

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

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

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

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

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

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

$d=2.1\ miles$

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What is the sum of 2/10 + 6/100

kogti [31]

Answer:

26/100 which is simplified to 13/50

Step-by-step explanation:

you have to get equal denominators so multiply 2/10 by 10 on top and bottom to get 20/100 then you can add 6/100 plus 20/100 which is 26/100 . denominators stay the same 26/100 can be simplified to 13/50

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2 years ago

Read 2 more answers

A rectangular portrait measures 50cm by 70cm. It is surrounded by a rectangular frame of uniform width. If the area of the frame

snow_tiger [21]

Let us assume uniform width = x cm wide.

Length of rectangular portrait = 50cm and width of rectangular portrait = 70cm.

Therefore, length of rectangle made by frame = 50 + x+x = (2x+50) cm.

And width of rectangle made by frame = 70+x+x = (2x+70) cm.

We know, the area of rectangular portrait = 50 × 70 = 3500 cm^2.

Total area of the rectangle made by frame would be = (2x+50) * (2x+70)

We know,

Actual area of frame = Area of rectangle made by frame - area of rectangular portrait.

We also given "the area of the frame is the same as the area of the portrait."

We can setup an equation now,

3500 = (2x+50) * (2x+70) - 3500.

Subtracting 3500 from both sides, we get

3500-3500 = (2x+50) * (2x+70) - 3500-3500.

0 = (2x+50) * (2x+70) -7000.

FOIL (2x+50) * (2x+70), we get

0 = 2x*2x +2x*70 + 50*2x +50*70 - 7000.

0 = 4x^2 +140x +100x +3500 -7000.

4x^2 +240x -3500 = 0.

Dividing whole equation by 4, we get

x^2 +60x - 875 =0

Applying quadratic formula $=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , we get

$=\frac{-60\pm \sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}$

$x=\frac{-60+\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}=5\left(\sqrt{71}-6\right)$

$x=\frac{-60-\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}:\quad -5\left(6+\sqrt{71}\right)$

We cant take negative value.

So, $x=5\left(\sqrt{71}-6\right)=12.13$

We could take approximately 12 cm.

7 0

2 years ago

APEX HELP ASAP!!!!

oee [108]

1. The two boats picked for the trip are the steamboat and the tall ship. Let us assume that we will take the steamboat going to the island, and then we will take the tall ship for the return trip. We will then relate the distances travelled by both ships to each other.

2. We know that the steamboat takes five hours to complete the trip. The tall ship takes more time, at ten hours to complete the trip. We do not have the exact speeds of the steamboat or of the tall ship, but we do know that the tall ship is 10 knots slower than the steamboat. We likewise do not know the exact distance travelled by either ship, but we do know that both travel the same distance. We want to find out how fast each boat travels. We expect the answers to be in knots, with a difference of 10.

3. We know that distance is equivalent to the product of speed of a boat multiplied by the time of travel. For the trip going to the island, we will use the steamboat. Let its speed be x knots (equivalent to x nautical miles per hour), and let the distance going to the island be d nautical miles. Given that the time takes is 5 hours, this means that d = 5x.

4. If we let x be the speed of the boat you are taking to the island (the steamboat), then we know that the speed of the other boat (the tall ship) is 10 knots less than the steamboat's. So the speed of the tall ship (for the return trip) is (x - 10) knots.

5. Similar to part 3: we will multiply speed by time to determine the distance from the island. From part 4, we have determined that the speed of the tall ship to be used in returning is (x - 10) knots. Meanwhile, the given in the problem says that the tall ship will take 10 hours to make the trip. Therefore the distance will be equal to d = 10(x - 10) = 10x - 100 nautical miles.

6. We can assume that the distance travelled going to the island is the same distance travelled coming back. Therefore, we can equate the formula for distance from part 3 for the steamboat, to the distance from part 5 for the tall ship.
5x = 10x - 100

7. Solving for x: 5x = 10x - 100
-5x = -100
x = 20
Since x is the speed of the steamboat, x = 20 means that the steamboat's speed is 20 knots.

8. We determined in part 4 that the speed of the second boat (in our case, the tall ship) is (x - 10) knots. Since we have calculated in part 7 that the steamboat travels at x = 20 knots, then the speed of the tall ship is (x - 10) = 20 - 10 = 10 knots.

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2 years ago

The minimum distance from Earth to the Sun is 91.4 million miles. The maximum distance is 94.5 million miles write a absolute va

d1i1m1o1n [39]

Solution :

The minimum distance from Earth to the Sun is 91.4 million miles

The maximum distance is 94.5 million miles

$d_{min} = 91.4, d_{max} = 94.5$