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3 years ago

14

At a warehouse, boxes of merchandise are placed on shelves in stacks that are 7 boxes high. If each box is 3 1/5 inches in heigh

t, how tall is the stack of boxes.

1 answer:

lesya692 [45]3 years ago

6 0

22.4; 7 * 3 1/5 = 22.4

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If a line is 10 feet long but needs to be reduced to 80% of its original, how long will the new line be?

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Answer:

8 feet

Step-by-step explanation:

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3 years ago

Suppose that the function

Dafna11 [192]

Answer:

<em>(</em><em>f-g</em><em>)</em><em>(</em><em>x</em><em>)</em><em>=</em><em> </em><em>-3x</em><em>^</em><em>3</em><em> </em><em>+</em><em> </em><em>x</em><em>^</em><em>2</em><em> </em><em>+</em><em> </em><em>6x</em>

<em>(</em><em>f</em><em>+</em><em>g</em><em>)</em><em>(</em><em>×</em><em>)</em><em>=</em><em> </em><em>3x</em><em>^</em><em>3</em><em> </em><em>+</em><em> </em><em>x</em><em>^</em><em>2</em><em> </em><em>+</em><em> </em><em>6x</em>

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Step-by-step explanation:

I hope that helps

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3 years ago

Look ate the picture down below, and please help me i would really appreciate it

zavuch27 [327]

Answer:

46°

Step-by-step explanation:

When secants intersect each other and a circle, the external angle (A) is half the difference of the intercepted arcs:

∠A = (arcDC -arcBC)/2

12° = (arcDC -22°)/2 . . . . . . . fill in the given numbers

24° = arcDC -22° . . . . . . . . . multiply by 2

46° = arcDC . . . . . . . . . . . . . add 22°

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3 years ago

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Answer:

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2 years ago

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.

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3 years ago