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Sergeu [11.5K]
3 years ago
12

Which method correctly solves the equation using the multiplication property of equality and the reciprocal of One-third? One-th

ird (x + 9) = negative 12 One-third (x + 9 = negative 12. One-third x + 3 = negative 4. One-third x = negative 7. x = negative 27.
Mathematics
2 answers:
LenKa [72]3 years ago
8 0

Answer:

Minus 12

Step-by-step explanation:

ITS CORRECT

Airida [17]3 years ago
4 0

Answer:

Step-by-step explanation:

Given the equation \frac{1}{3}(x+9) = -12

Step 1;

Expand the bracket at the right hand side of the equation to have:

\frac{1}{3}x +\frac{1}{3}(9) = -12\\\frac{1}{3}x+3=-12\\subtracting\ 3\ from\ both\ sides\\\frac{1}{3}x+3-3=-12-3\\\frac{1}{3}x=-15\\

Taking the reciprocal of both sides:

\frac{3}{x} = \frac{-1}{15} \\ cross\ multiplying\\-x =45\\x=-45

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I really need help in math! Any is appreciated!
Usimov [2.4K]
The answer should be 7/5
5 0
3 years ago
Suppose you could climb to the top of the Great Pyramid of Giza in Egypt. Which path would be shorter, climbing a lateral edge o
valentinak56 [21]
Check the picture below.

if you go from the slant height, the path is shorter to the top, since the hypotenuse is the longest of all sides in a right-triangle.

that said, is shorter but is also steeper, so the lateral edge is longer but less steep and thus a bit simpler though longer, but we're only asked on which is shorter.

5 0
3 years ago
the area of a window is 40 square feet. the length is the square root of 20 feet. what is the width of the window
Irina18 [472]
----------------------------------------------------
Given information
----------------------------------------------------
Area = 40ft²
Length = √20 ft

----------------------------------------------------
Formula
----------------------------------------------------
Area = Length x Width

----------------------------------------------------
Find Width
----------------------------------------------------
40 = √20 x width
Width = 40 ÷ √20
Width = 40 ÷ 2√5
Width = 20 ÷ √5
Width =  \frac{20}{ \sqrt{5} }

Answer: \boxed {Width = \frac{20}{ \sqrt{5} }}

3 0
3 years ago
A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot
ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box  is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = (x\times x)

                               =x^2

The volume of the box  is = area of the base × height

                                           =x^2h

Therefore,

x^2h=272

\Rightarrow h=\frac{272}{x^2}

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

=20x^2 cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

=10x^2 cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

=4xh\times 1.5 cents

=6xh cents

Total cost = (20x^2+10x^2+6xh)

                =30x^2+6xh

Let

C=30x^2+6xh

Putting the value of h

C=30x^2+6x\times \frac{272}{x^2}

\Rightarrow C=30x^2+\frac{1632}{x}

Differentiating with respect to x

C'=60x-\frac{1632}{x^2}

Again differentiating with respect to x

C''=60+\frac{3264}{x^3}

Now set C'=0

60x-\frac{1632}{x^2}=0

\Rightarrow 60x=\frac{1632}{x^2}

\Rightarrow x^3=\frac{1632}{60}

\Rightarrow x\approx 3

Now C''|_{x=3}=60+\frac{3264}{3^3}>0

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is =\frac{272}{3^2}

                                          =30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

7 0
3 years ago
What is the answer??????????
Troyanec [42]
The answer is 91 ............
4 0
2 years ago
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