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pickupchik [31]
2 years ago
13

Can someone please help me on these questions?

Mathematics
2 answers:
Lilit [14]2 years ago
8 0
A is equal to 4 3/5
And b is equal to 3 1/6
sleet_krkn [62]2 years ago
7 0
The first one is equal to 4 3/5


Because we add the front and add te numerator for the battle its stay the same
The second one is 3 1/1 so it's 3 1

We jut subtrat
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Explain how you can finding 4 384
kkurt [141]
There are a lot of ways you can do this, depending on what numbers you use, you can divide, multiply, add, or subtract to get 4,384, did you mean a specific term of math?<span />
8 0
2 years ago
Read 2 more answers
I will give brainliest!! please help!!!
hram777 [196]

Answer:

Yes

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6 0
3 years ago
Suppose you buy 4 buckets of apples and 5 buckets of peaches for $64. If you bought 8
Sophie [7]

Option D is the correct answer.

Step-by-step explanation:

Step 1 :

Let A represent the cost of one bucket of apple and P represent the cost of one bucket of peaches.

So we have ,

4 buckets of apples and 5 buckets of peaches for $64

8  buckets of apples and 3 buckets of peaches for $72

Writing this in the equation form we have,

4A + 5P = 64

8A + 3P = 72

Step 2:

Solving for the above 2 equations we can get the required costs

Equation 1 is

4A + 5P = 64 , Multiplying this by 2 we have 8A + 10P = 128

Equation 2  = 8A + 3P = 72

Subtracting both we have , 7p = 56 = > P = 8

Substituting this in equation 1 we have

8A + 80 = 128 => A = 6

Hence the cost of one bucket of apple is $6 and the cost of one bucket of apple is $8.

Option D is the correct answer.

3 0
3 years ago
Read 2 more answers
How much time did Cole spend painting?
Katyanochek1 [597]
1 and 11/12 hours I think
6 0
3 years ago
Solve for x and y.<br><br> 4x-3y=16<br> -2x+4y=-2<br><br> DUE 2/7/17!
Trava [24]
\left \{ {{4x-3y=16\:(I)} \atop {-2x+4y=-2\:(II)}} \right.

<span>Multiply the 2nd equation by (-4). This will eliminate the x's when you add the two new equations together.
</span>\left \{ {{4x-3y=16\:\:\:\:\:\:\:\:\:\:} \atop {-2x+4y=-2\:*(-4)}} \right.
\left \{ {{\diagup\!\!\!\!4x-3y=16} \atop {-\diagup\!\!\!\!4x+8y=-4}} \right.
--------------------
5y = 12
\boxed{\boxed{y =  \frac{12}{5}}}\end{array}}\qquad\quad\checkmark

Let's replace the value found in the first equation:
4x-3y=16\:(I)
4x-3*( \frac{12}{5}) =16
4x -  \frac{36}{5} = 16
least common multiple (5)
\frac{20x}{\diagup\!\!\!\!5} - \frac{36}{\diagup\!\!\!\!5} = \frac{80}{\diagup\!\!\!\!5}
20x - 36 = 80
20x = 80 + 36
20x = 116
x = \frac{116}{20} \frac{\div4}{\div4} \to\: \boxed{\boxed{x = \frac{29}{5} }}\end{array}}\qquad\quad\checkmark



4 0
3 years ago
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