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3 years ago

*If you add two integers with different signs, how can you tell whether the sum will be positive, negative, or zero?

Mathematics

1 answer:

irina [24]3 years ago

7 0

Answer:

Please see explanation

Step-by-step explanation:

When two integers are being added there can be three cases in the answer

Positive

The sum will be positive if both the integers are positive and in case of addition of one positive and negative integer, the sum will be positive if the integer with larger magnitude is positive

Negative

The sum will be negative if both integers are negative and in case of addition of one positive and negative integer, the sum will be negative if the integer with larger magnitude is negative

Zero

The sum will be zero if both the integers have same magnitude and different signs.

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Which expression represents the calculation “add 5 and 6, then multiply by 3"?

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Answer:

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3 years ago

– Jack can paint a room in 9 hours alone. If Sara helps, they can do the job in 5 hours. How long

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Answer: 9

Step-by-step explanation:

If Jack can paint the room in 9 hours by his self she can only do the same

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3 years ago

one canned juice drink is 25% orange juice ; another is 10% orange juice. how many liters of each should be mixed towing order t

JulijaS [17]

-------------------------------------------------------------------------------------------------
Assumptions
-------------------------------------------------------------------------------------------------
Let x be the amount of 25% orange juice
Let y be the amount of 10% orange juice

-------------------------------------------------------------------------------------------------
There are 15 litres of 13% orange juice
-------------------------------------------------------------------------------------------------
x + y = 15

-------------------------------------------------------------------------------------------------
The 15 litres of juice is 13% orange juice
-------------------------------------------------------------------------------------------------
0.25x + 0.1y = 15 x 0.13
0.25x + 0.1y = 1.95

-------------------------------------------------------------------------------------------------
Rewrite the equations and solve for x and y
-------------------------------------------------------------------------------------------------
x + y = 15 ----------------------- (Equation 1)
0.25x + 0.1y = 1.95 ----------- (Equation 2)

(Equation 1):
x + y = 15
x = 15 - y ----------- (Substitute into Equation 2)

0.25 (15 - y) + 0.1y = 1.95
3.75 - 0.25y + 0.1y = 1.95
0.15y = 1.8
y = 18 ÷ 0.15
y = 12 ------------- (substitute into Equation 1)

x + y = 15
x + 12 = 15
x = 15 - 12
x = 3

x = 3 and y = 12

-------------------------------------------------------------------------------------------------
Answer: 3 liters of 25% of orange juice with 12 liters of 10% orange juice
-------------------------------------------------------------------------------------------------

7 0

3 years ago

Mrs. Parson baked 25 cookies. 4 fifths of the cookies were oatmeal. How many oatmeal cookies did Mrs. Parson bake?

lutik1710 [3]

Answer:

I think the answer is 20 ..........

Step-by-step explanation:

7 0

3 years ago

Help me to answer now ineed this 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