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yarga [219]
3 years ago
13

How do you add and subtract negative mixed numbers and fractions?

Mathematics
1 answer:
horsena [70]3 years ago
3 0
Well, there are a few ways but you do it like you would integers. For instance, -6 + 2 = -4.  -\frac{3}{5} + \frac{1}{5} = -\frac{2}{5} Also if you have fraction with different denominators you would first make the fractions with the same denominator and then add and subtract like you would integers like
-4 - 6 = -10


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The graph shows the relationship between the number of pounds of
Damm [24]

Given:

The graph shows the relationship between the number of pounds of  apples purchased and the total cost of the apples.

To find:

The cost of 1 pound of apples.

Solution:

We need to find the slope of the line to get the cost of 1 pound of apples.

From the given graph it is clear that the line passes through two points (0,0) and (2,3). So, the slope of the line is

m=\dfrac{y_2-y_1}{x_2-x_1}

m=\dfrac{3-0}{2-0}

m=\dfrac{3}{2}

m=1.5

Therefore, the cost of 1 pound of apples is $1.5.

3 0
3 years ago
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Juan ran the lemonade stand for 3 more days. Each day, he used the money from sales to purchase more lemons, cups, and sugar to
pav-90 [236]

Answer:

c.

Step-by-step explanation:

7 0
4 years ago
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And another! Apparently you use the distributive property to remove the parentheses. .-.
Tju [1.3M]
6v+4w-4
Because you do -*-6v=positive 6v then -*-4w=positive 4w and then -*4=-4 and that takes care of the parenthesis
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8 0
3 years ago
400,000+60,000+5,000+100 in word form
lord [1]
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3 0
3 years ago
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BoxedNGone truck rentals calculates that its price function is p(x) = 200 − 2x, where p is the price (in dollars) at which exact
charle [14.2K]

Answer:

At Maximum point;

x(max) = 50

p(max) = $100

Maximum revenue = $5,000

Step-by-step explanation:

The price function is;

p(x) = 200 − 2x

where

p is the price (in dollars) at which exactly x trucks will be rented per day.

The revenue function R(x) can be written as;

R(x) = p(x) × x

Substituting p(x) equation;

R(x) = (200-2x)x

R(x) = 200x-2x^2 ........1

To maximize R(x), at maximum point dR/dx = 0

differentiating equation 1;

dR/dx = 200 - 4x = 0

4x = 200

x = 200/4

x = 50

Substituting x = 50 into p(x)

p(50) = 200 - 2(50) = $100

p = $100

Maximum revenue is;

R = p × x = $100×50

R = $5,000

3 0
3 years ago
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