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2 years ago

5

PLSS HELP Sarah left the boat dock and sailed 5 miles due east. She turned and then sailed 10 miles due north. About how far is

Sarah from the boat dock?
Group of answer choices
10 miles

9 miles

15 miles

11 miles

1 answer:

Sever21 [200]2 years ago

7 0

Answer:

I believe the answer is 11 miles

Step-by-step explanation:

I drew the problem out and used the Pythagorean theorem and got 11.2

Hope this helps you!

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A rectangle has a length 6 more than it's width if the width is decreased by 2 and the length decreased by 4 the resulting has a

Rashid [163]

Answer:

Length of original rectangle: 11 units.

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

$\text{Perimeter of new rectangle}=20$

Step-by-step explanation:

Let x represent width of the original rectangle.

We have been given that a rectangle has a length 6 more than it's width. S the length of the original rectangle would be $x+6$ .

We have been given that when the width is decreased by 2 and the length decreased by 4 the resulting has an area of 21 square units.

The width of new rectangle would be $x-2$ .

The length of new rectangle would be $x+6-4=x+2$ .

The area of new rectangle would be $(x+2)(x-2)$ .

Now we will equate area of new rectangle with 21 and solve for x as:

$(x+2)(x-2)=21$

Applying difference of squares, we will get:

$x^2-2^2=21$

$x^2-4=21$

$x^2-4+4=21+4$

$x^2=25$

Since width cannot be negative, so we will take positive square root of both sides.

$\sqrt{x^2}=\sqrt{25}$

$x=5$

Therefore, the width of original rectangle is 5 units.

Length of the original rectangle would be $x+6\Rightarrow x+5=11$ .

Therefore, the length of original rectangle is 11 units.

$\text{Area of original rectangle}=5\times 11$

$\text{Area of original rectangle}=55$

Therefore, area of the original rectangle is 55 square units.

Now we will find ratio of the original rectangle area to the new rectangle area as:

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

We know that perimeter of rectangle is two times the sum of length and width.

$\text{Perimeter of new rectangle}=2((x+2)+(x-2))$

$\text{Perimeter of new rectangle}=2((5+2)+(5-2))$

$\text{Perimeter of new rectangle}=2(7+3)$

$\text{Perimeter of new rectangle}=2(10)$

$\text{Perimeter of new rectangle}=20$

Therefore, the perimeter of the new rectangle is 20 units.

7 0

3 years ago

A metallurgist has one alloy containing 32% copper and another containing 65% copper. How many pounds of each alloy must he use

kompoz [17]

Answer:

<h2>30.67 pounds of 32% copper alloy and 10.33 pounds of 65% copper alloy</h2>

Step-by-step explanation:

First alloy contains 32% copper and the second alloy contains 65% alloy.

We wish to make 44 pounds of a third alloy containing 42% copper.

Let the weight of first alloy used be $x$ in pounds and the weight of second alloy used be $y$ in pounds.

Total weight = $44\text{ }pounds=x+y$ $-(i)$

Total weight of copper = $42\%\text{ of 44 pounds = }32\%\text{ of }x\text{ pounds + }65\%\text{ of }y\text{ pounds }$

$\dfrac{42\times 44}{100}=\dfrac{32x}{100}+\dfrac{65y}{100}\\\\ 32x+65y=1848$ $-(ii)$

Subtracting 32 times first equation from second equation,

$32x+65y-32x-32y=1848-32\times44\\33y=440\\y=13.333\text{ }pounds \\x=30.667\text{ }pounds$

∴ 30.67 pounds of first alloy and 13.33 pounds of second alloy were used.

4 0

2 years ago