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3 years ago

How do you simplify 20/36

2 answers:

8 0

$\frac { 20 }{ 36 } \\ \\ =\frac { 5 }{ 9 } \cdot \frac { 4 }{ 4 } \\ \\ =\frac { 5 }{ 9 } \cdot 1\\ \\ =\frac { 5 }{ 9 }$

horsena [70]3 years ago

5 0

$\frac{20}{36} \\ 
 \\ \frac{5}{9} \ x \ \frac{4}{4} \ = \frac{20}{36}
 \\ \frac{4}{4} \ = \ 1
 \\ \frac{5}{9} \ x \ 1 \ = \ \frac{5}{9}
 \\ Solution: \frac{5}{9} \\ 
$

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A photo is 8 cm long and 6 cm wide. It is enlarged to make a print that is 18 cm wide.

dezoksy [38]

Answer:

24 cm

Step-by-step explanation:

The photo was 6 cm wide, and was made into 18 cm wide. Divide 18 with 6:

18/6 = 3

To solve for the length enlargement, multiply 3 with the length, or 8 cm:

8 x 3 = 24

24 cm is the length of the enlargement.

3 0

3 years ago

Jack wants to share his candy with friends. Right now he has 4 1/3 pieces of chocolate. How many friends can he give 1/2 a candy

cricket20 [7]

Answer:

8 Friends

Step-by-step explanation:

First convert 4 1/3 to a mixed fraction

4 1/3 --> 13/3

Then divide by 1/2 to find how many friends per 1/2 candy bar with 13/3 candy bars

(13/3)/(1/2)

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4 years ago

x can do a job in 16 days and y can do the same jovb in 20 days x y and z can do the job in 7.5 days. alone z can do the same jo

MissTica

Answer:

24 days

Step-by-step explanation:

4 0

4 years ago

Read 2 more answers

Please help me, A candle is in the shape of a frustum (a cone with the top cut off). Calculate the volume of the candle to the n

MariettaO [177]

Answer:

Height = 8 cm

You will need this formula (see attached):

Volume = [PI * 8 * (2² + 2*4 + 4²)] / 3

Volume = [3.14159265 * 8 * (4 + 8 + 16)] / 3

Volume = 234.57 cubic centimeters

Source: http://www.1728.org/volcone.htm

Step-by-step explanation:

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2 years ago

Consider the given matrix. 1 −1 0 5 1 4 0 1 1 Find the eigenvalues. (Enter your answers as a comma-separated list.) λ = Find the

fomenos

Answer:

Step-by-step explanation:

We have to diagonalize the matrix

$\left[\begin{array}{ccc}1&-1&0\\5&1&4\\0&1&1\end{array}\right]$

we have to solve the expression

$|A-\lambda I|=0$

Thus, by applying the determinant we obtain the polynomial

$(1-\lambda )^3+5-4=0\\(1-\lambda )^3+1=0\\$

$-\lambda^3+3\lambda^2-4\lambda+2$

$\lambda_1=1\\\lambda_2=1-i\\\lambda_3=1+i\\$

and the eigenvector will be

$v_1=(-4,0,5)\\v_2=(-1,-i,-1)\\v_3=(-1,i,1)$

HOPE THIS HELPS!!

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3 years ago