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Zina [86]
2 years ago
14

Four ninths plus one ninth

Mathematics
2 answers:
marshall27 [118]2 years ago
8 0

Answer:

5/9

Step-by-step explanation:

SIZIF [17.4K]2 years ago
7 0

Answer:

5/9

Step-by-step explanation:

4/9 + 1/9

To add fractions, you need a common denominator. You already have a common denominator of 9, so just add the numerators.

4/9 + 1/9 = 5/9

Since there is no common factor of 5 and 9 other than 1, the fraction is already reduced.

Answer: 5/9

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A smaller square of side length 17 feet is cut out of a square board. What is the approximate area (shaded region) of the remain
gregori [183]

Answer:

The area of the remaining board is (<em>x</em>² - 289) sq. ft.

Step-by-step explanation:

Let the sides of the bigger square board be, <em>x</em> feet.

It is provided that a smaller square of side length 17 feet is cut out of the bigger square board.

The area of a square is:

Area=(side)^{2}

Compute the area of the bigger square board as follows:

A_{b}=(side_{b})^{2}=x^{2}

Compute the area of the smaller square board as follows:

A_{s}=(side_{s})^{2}=(17)^{2}=289

Compute the area of the remaining board in square feet as follows:

\text{Remaining Area}=A_{b}-A_{s}

                          =[x^{2}-289]\ \text{square ft.}

Thus, the area of the remaining board is (<em>x</em>² - 289) sq. ft.

7 0
3 years ago
What is the answer4-2.1(3.4-1.6)
tensa zangetsu [6.8K]

Answer: 0.22 I hope i helped

Steps: 3.14 - 1.6 = 1.8

1.8 x 2.1 = 3.78

3.78 - 4 = 0.22



8 0
3 years ago
Read 2 more answers
A construction crew has just finished building a road. The crew worked for 2 7/9 days. If they built 2 1/5 kilometers of road ea
IRISSAK [1]

let 'x' represent the number of kilometers built per day


(5 2/5)x = 12


((5*5 + 2)/5)x = 12


by solving we find


x = 2.22 km  hope this is right 

5 0
3 years ago
Pls answer this and I will mark it has brainlist j
mafiozo [28]

Answer: C=250+4s

The reason is because the cost is equal to $250, which is the bus trip, plus $4 additional dollars for each student, s.

4 0
3 years ago
Verify that each equation is an identity (1 - sin^(2)((x)/(2)))/(1+sin^(2)((x)/(2)))= (1+cosx)/(3-cosX)
Allisa [31]

Answer:

Given that we have;

sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }

By the application of the law of indices and algebraic process of adding a and subtracting a fraction from a whole number, we have;

\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) }  =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

Step-by-step explanation:

An identity is a valid or true equation for all variable values

The given equation is presented as follows;

\dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

From trigonometric identities, we have;

sin \left (\dfrac{x}{2} \right ) = \sqrt{\dfrac{1 - cos (x)}{2} }

\therefore sin^2 \left (\dfrac{x}{2} \right ) = \dfrac{1 - cos (x)}{2}

1 -  sin^2 \left (\dfrac{x}{2} \right ) = 1 - \dfrac{1 - cos (x)}{2} = \dfrac{2 - (1 - cos (x))}{2} = \dfrac{1 + cos (x))}{2}

1 +  sin^2 \left (\dfrac{x}{2} \right ) = 1 + \dfrac{1 - cos (x)}{2} = \dfrac{2 + 1 - cos (x))}{2} = \dfrac{3 - cos (x))}{2}

\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( \dfrac{1 + cos (x)}{2} \right)}{\left (\dfrac{3 - cos (x)}{2} \right ) }  =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

\therefore \dfrac{\left ( 1 - sin^2 \left (\dfrac{x}{2} \right ) \right )}{\left ( 1 + sin^2 \left (\dfrac{x}{2} \right ) \right )} =\dfrac{\left ( 1 + cos (x))}{(3 - cos (x))}

3 0
3 years ago
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