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Ivan
2 years ago
7

After Paul hiked 5/6 of a mile, he was 2/3 of the way along the trail. How long is the trail?

Mathematics
1 answer:
Anvisha [2.4K]2 years ago
3 0

Answer:

1.25 miles.

Step-by-step explanation:

Let x be the length of trail. We have been given that after Paul hiked \frac{5}{6} of a mile, he was \frac{5}{6} of the way along the trail.  

Let need to find x such that \frac{2}{3} of x equals \frac{5}{6}.  

\frac{2}{3} x=\frac{5}{6}

x=\frac{5}{6}\times \frac{3}{2}

x=\frac{5}{2}\times\frac{1}{2}  

x=\frac{5}{2\times 2}

x=\frac{5}{4}  

x=1\frac{1}{4}=1.25  

Therefore, the trail is 1.25 miles long.  





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Two boats leave port at noon. Boat 1 sails due east at 12 knots. Boat 2 sails due south at 8 knots. At 2 pm the wind diminishes
Ivan

Answer:

14.86 knots.

Step-by-step explanation:

<em>Given that:</em>

The boats leave the port at noon.

Speed of boat 1 = 12 knots due east

Speed of boat 2 = 8 knots due south

At 2 pm:

Distance traveled by boat 1 = 24 units due east

Distance traveled by boat 2 = 16 units due south

Now, speed of boat 1 changes to 9 knots:

At 3 pm:

Distance traveled by boat 1 = 24 + 9= 33 units due east

Distance traveled by boat 2 = 16+8 = 24 units due south

Now, speed of boat 1 changes to 8+7 = 15 knots

At 5 pm:

Distance traveled by boat 1 = 33 + 2\times 9= 51 units due east

Distance traveled by boat 2 = 24 + 2 \times 15 = 54 units due south

Now, the situation of distance traveled can be seen by the attached right angled \triangle AOB.

O is the port and A is the location of boat 1

B is the location of boat 2.

Using pythagorean theorem:

\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow AB^{2} = OA^{2} + OB^{2}\\\Rightarrow AB^{2} = 51^{2} + 54^{2}\\\Rightarrow AB^{2} = 2601+ 2916 = 5517\\\Rightarrow AB = 74.28\ units

so, the total distance between the two boats is 74.28 units.

Change in distance per hour = \dfrac{Total\ distance}{Total\ time}

\Rightarrow \dfrac{74.28}{5} = 14.86\ knots

6 0
2 years ago
Given the image below, what is the image of C after a dilation with a scale factor of 3
Ne4ueva [31]

Answer: The coordinates of point C after the dilation are (-2, 5)

Step-by-step explanation:

I guess that you want to find where the point C ends after the dilation.

Ok, if we have a point (x, y) and we do a dilation with a scale A around the point (a,b), then the dilated point will be:

(a + A*(x - a), b + A*(y - b))

In this case we have:

(a,b) = (2,1) and A = 3.

And the coordinates of point C, before being dilated, are: (1, 2)

Then the new location of the point C will be:

C' = (1 + 3*(1 - 2), 2 + 3*(2 - 1)) = (1 -3, 2 + 3) = (-2, 5)

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2 years ago
Hailey is shopping at a department store during a 20% off everything sale. She also has a coupon for $5.00 off the sale amount.
Kipish [7]

Possible inequality:

= x\leq 87.50

4 0
2 years ago
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An oblong box has a volume equal to lwh, where l is the length, w is the width, and h is the height. If the volume is 24 cubic f
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<span>An oblong box has a volume equal to lwh, where l is the length, w is the width, and h is the height. If the volume is 24 cubic feet, solve for the height in terms of the other sides.

Given:
volume of 24 cubic feet

Required:
height

Solution:
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2 years ago
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weeeeeb [17]

Answer:

20,475\ ways

Step-by-step explanation:

we know that

<u><em>Combinations</em></u> are a way to calculate the total outcomes of an event where order of the outcomes does not matter.

To calculate combinations, we will use the formula

C(n,r)=\frac{n!}{r!(n-r)!}

where

n represents the total number of items

r represents the number of items being chosen at a time.

In this problem

n=28\\r=4

substitute

C(28,4)=\frac{28!}{4!(28-4)!}\\\\C(28,4)=\frac{28!}{4!(24)!}

simplify

C(28,4)=\frac{(28)(27)(26)(25)(24!)}{4!(24)!}

C(28,4)=\frac{(28)(27)(26)(25)}{(4)(3)(2)(1)}

C(28,4)=20,475\ ways

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2 years ago
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