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3 years ago

How do you split $11 among 2 people

2 answers:

7 0

You divide 11 by 2 once you get your answer round it to the nearest tenth so when you divide 11 and 2 you get 5.5 you round it and your answer is 6

Fofino [41]3 years ago

5 0

U either don't and keep the money all to your self or 5 1/2 dollars to each person

You might be interested in

The formula below is used to convert a temperature in dregrees Celsius , C , to a temperature in dregrees farenheit , F .

lesya [120]

1.8(15)+32
1.8(15)=27
27+32+59

3 0

3 years ago

HELPPL Enter the solution from least to greatest (x-2)(3x+3)=0

emmainna [20.7K]

Answer:

x = -1, 2

Step-by-step explanation:

Step 1: Write out equation

(x - 2)(3x + 3) = 0

Step 2: Solve each individual root

x - 2 = 0

x = 2

3x + 3 = 0

3x = -3

x = -1

Step 3: Arrange from least to greatest

-1 < 2

So x = -1, 2

7 0

3 years ago

Read 2 more answers

Please help ill mark brainliest

sp2606 [1]

Answer: She is correct because the average of the numbers 2, 3, 1, 4, 5, 2, 3, 4, 3, is 3.

Step-by-step explanation:

To find average you add all of the listed numbers, (2, 3, 1, 4, 5, 2, 3, 4, 3) which is 27. You then divide 27 by the amount of listed numbers, which is 9. 27 divided by 9 is 3. There for the average amount of strokes it takes to hit the ball into the hole of each green is 3.

Hope this helps! Please rate, it helps me know if I am explaining the answers well enough or not!! Thank you!<3

7 0

3 years ago

Help me to answer now ineed this <br> Please...

Vera_Pavlovna [14]

ANSWER TO QUESTION 1

$\frac{\frac{y^2-4}{x^2-9}} {\frac{y-2}{x+3}}$

Let us change middle bar to division sign.

$\frac{y^2-4}{x^2-9}\div \frac{y-2}{x+3}$

We now multiply with the reciprocal of the second fraction

$\frac{y^2-4}{x^2-9}\times \frac{x+3}{y-2}$

We factor the first fraction using difference of two squares.

$\frac{(y-2)(y+2)}{(x-3)(x+3)}\times \frac{x+3}{y-2}$

We cancel common factors.

$\frac{(y+2)}{(x-3)}\times \frac{1}{1}$

This simplifies to

$\frac{(y+2)}{(x-3)}$

ANSWER TO QUESTION 2

$\frac{1+\frac{1}{x}} {\frac{2}{x+3}-\frac{1}{x+2}}$

We change the middle bar to the division sign

$(1+\frac{1}{x}) \div (\frac{2}{x+3}-\frac{1}{x+2})$

We collect LCM to obtain

$(\frac{x+1}{x})\div \frac{2(x+2)-1(x+3)}{(x+3)(x+2)}$

We expand and simplify to obtain,

$(\frac{x+1}{x})\div \frac{2x+4-x-3}{(x+3)(x+2)}$

$(\frac{x+1}{x})\div \frac{x+1}{(x+3)(x+2)}$

We now multiply with the reciprocal,

$(\frac{(x+1)}{x})\times \frac{(x+2)(x+3)}{(x+1)}$

We cancel out common factors to obtain;

$(\frac{1}{x})\times \frac{(x+2)(x+3)}{1}$

This simplifies to;

$\frac{(x+2)(x+3)}{x}$

ANSWER TO QUESTION 3

$\frac{\frac{a-b}{a+b}} {\frac{a+b}{a-b}}$

We rewrite the above expression to obtain;

$\frac{a-b}{a+b}\div {\frac{a+b}{a-b}}$

We now multiply by the reciprocal,

$\frac{a-b}{a+b}\times {\frac{a-b}{a+b}}$

We multiply out to get,

$\frac{(a-b)^2}{(a+b)^2}$

ANSWER T0 QUESTION 4

To solve the equation,

$\frac{m}{m+1} +\frac{5}{m-1} =1$

We multiply through by the LCM of $(m+1)(m-1)$

$(m+1)(m-1) \times \frac{m}{m+1} + (m+1)(m-1) \times \frac{5}{m-1} =(m+1)(m-1) \times 1$

This gives us,

$(m-1) \times m + (m+1) \times 5}=(m+1)(m-1)$

$m^2-m+ 5m+5=m^2-1$

This simplifies to;

$4m-5=-1$

$4m=-1-5$

$4m=-6$

$\Rightarrow m=-\frac{6}{4}$

$\Rightarrow m=-\frac{3}{2}$

ANSWER TO QUESTION 5

$\frac{3}{5x}+ \frac{7}{2x}=1$

We multiply through with the LCM of $10x$

$10x \times \frac{3}{5x}+10x \times \frac{7}{2x}=10x \times1$

We simplify to get,

$2 \times 3+5 \times 7=10x$

$6+35=10x$

$41=10x$

$x=\frac{41}{10}$

$x=4\frac{1}{10}$

Method 1: Simplifying the expression as it is.

$\frac{\frac{3}{4}+\frac{1}{5}}{\frac{5}{8}+\frac{3}{10}}$

We find the LCM of the fractions in the numerator and those in the denominator separately.

$\frac{\frac{5\times 3+ 4\times 1}{20}}{\frac{(5\times 5+3\times 4)}{40}}$

We simplify further to get,

$\frac{\frac{15+ 4}{20}}{\frac{25+12}{40}}$

$\frac{\frac{19}{20}}{\frac{37}{40}}$

With this method numerator divides(cancels) numerator and denominator divides (cancels) denominator

$\frac{\frac{19}{1}}{\frac{37}{2}}$

Also, a denominator in the denominator multiplies a numerator in the numerator of the original fraction while a numerator in the denominator multiplies a denominator in the numerator of the original fraction.

That is;

$\frac{19\times 2}{1\times 37}$

This simplifies to

$\frac{38}{37}$

Method 2: Changing the middle bar to a normal division sign.

$(\frac{3}{4}+\frac{1}{5})\div (\frac{5}{8}+\frac{3}{10})$

We find the LCM of the fractions in the numerator and those in the denominator separately.

$(\frac{5\times 3+ 4\times 1}{20})\div (\frac{(5\times 5+3\times 4)}{40})$

We simplify further to get,

$(\frac{15+ 4}{20})\div (\frac{(25+12)}{40})$

$\frac{19}{20}\div \frac{(37)}{40}$

We now multiply by the reciprocal,

$\frac{19}{20}\times \frac{40}{37}$

$\frac{19}{1}\times \frac{2}{37}$

$\frac{38}{37}$

5 0

3 years ago

Signs are placed at the beginning and at the end of a 4-kilometer hiking trail. Signs are also placed every 500 meters along th

il63 [147K]

Answer:

9 Signs

Step-by-step explanation:

The hiking trail is 4 kilometer long.

Signs are placed at the beginning and at the end of the trail.

4km = 4000m

We can solve this using simple arithmetic progression.

The first term,a= 0
The common difference,d= 500.
The last term,l= 4000.

Therefore:

Last term, l=a+(n-1)d

4000=0+500(n-1)

4000=500n-500

4000+500=500n

4500=500n

Divide both sides by 500

n=9

Therefore, there are a total of 9 signs on the entire trail.

CHECK

The signs are placed at this marks

0m,500m,1000m,1500m,2000m,2500m,3000m,3500m and 4000m

8 0

4 years ago