Help with Pythagorean theorem

denis-greek [22]

Answer:

$\frac{7}{10} |$

Step-by-step explanation:

STEP 1:

2/3 + 7/10 = ?

The fractions have unlike denominators. First, find the Least Common Denominator and rewrite the fractions with the common denominator.

LCD(2/3, 7/10) = 30

Multiply both the numerator and denominator of each fraction by the number that makes its denominator equal the LCD. This is basically multiplying each fraction by 1.

$(\frac{2}{3}$ * $\frac{10}{10})$ + $(\frac{7}{10} * \frac{3}{3})$ = ?

Complete the multiplication and the equation becomes

$\frac{20}{30} + \frac{21}{30}$

The two fractions now have like denominators so you can add the numerators.

Then:

$\frac{20+21}{30} = \frac{41}{30}$

This fraction cannot be reduced.

The fraction 41/30

is the same as

41 divided by 30

Convert to a mixed number using

long division for 41 ÷ 30 = 1R11, so

41/30 = 1 11/30

Therefore:

2/3+7/10= 1 11/30

STEP 2:

41/30 + -2/3

The fractions have unlike denominators. First, find the Least Common Denominator and rewrite the fractions with the common denominator.

LCD(41/30, -2/3) = 30

Multiply both the numerator and denominator of each fraction by the number that makes its denominator equal the LCD. This is basically multiplying each fraction by 1.

$(\frac{41}{30} *\frac{1}{1} ) + ( \frac{-2}{3} * \frac{10}{10} )$

The two fractions now have like denominators so you can add the numerators.

Then:

$\frac{41+-20}{30} = \frac{21}{30}$

This fraction can be reduced by dividing both the numerator and denominator by the Greatest Common Factor of 21 and 30 using

GCF(21,30) = 3

$\frac{21/3}{30/3} =\frac{7}{10}$

Therefore:

$\frac{41}{30} + \frac{-2}{3} =\frac{7}{10}$ |

8 0

2 years ago

I need help with this please....Thanks

IrinaK [193]

Answer:

The total cost of redecoration is £66, that is less than £75. Tenant can decorate the room for the budget set.

Step-by-step explanation:

Lets start calculating the areas of shorter and longer walls.

The area of shorter wall is

2.65m*2.98m=7.897m^2.

The area of longer wall is

3.94m*2,65m=10.441m^2.

Together the area of one long and one short wall is

7.897m^2+10.441m^2=18.338m^2.

The area to be painted ( one short and one long wall) is 18.338m^2 that is less than 30m^2, and hence only one tin of paint is needed (costs £16).

If there is assumed that the sheets of wallpaper must be whole from floor to ceiling, the calculations are following.

We get from one roll

10m/2.65m=3.77 sheets; we can use only 3 of them

The width of the walls is 3,94m+2,98m=6,92m,

and division gives 6.92m/0.52m=13,3... .

Hence it requires 14 sheets, that is 5 rolls of wallpaper.

Wallpaper costs 5*£10=£50 and altogether it is needed £50+£16=£66 to redecorate the room.

7 0

3 years ago

Three students, Matthew, Thompson and Amos shared #15000 in the ratio of 5:2:x respectively. If Matthew's share is 7500, calcula

jasenka [17]

Answer:

The value of x is 3.

Step-by-step explanation:

Matthew earned 7500 of 15000, which is exactly half the total amount, meaning that the fraction corresponding to his earnings is 1/2.

Rate of 5:2:x, with number 5 corresponding to Matthew, mean that his fraction of the earnings, in function of x is given by:

$\frac{5}{5 + 2 + x} = \frac{5}{7 + x}$

Both fractions are equal, so:

$\frac{5}{7 + x} = \frac{1}{2}$

Applying cross multiplication:

$7 + x = 10$

$x = 10 - 7$

$x = 3$

The value of x is 3.

5 0

2 years ago

Alejandro has gone to school StartFraction 5 Over 7 EndFraction of the last 35 days. Which expression can be used to determine t

Arte-miy333 [17]

Answer:

Expression B

Step-by-step explanation:

Let us find the information in the question

Alejandro has gon to school $\frac{5}{7}$ of the last 35 days
We need to find the expression that represents the number of days Alejandro has gone to school.