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andrew11 [14]
2 years ago
12

A farm had 480 acres more wheat than corn. After the farmers collected 80% of the wheat and 25% of the corn, the area of the whe

at was 300 acres less than the area of the corn. What was the area of the wheat field and what was the area of the corn field?
Mathematics
1 answer:
Sergio [31]2 years ago
7 0
The area covered by corn is x
The area covered by wheat is (x+480)
After 80% of wheat was collected, the new area was:
20/10(x+480)
=1/5(x+480)
After 25% of the corn was collected, the new area was:
75/100x
=3/4x
New area equation will be:
(3/4)x=1/5(x+480)+300
(3/4x)=(1/5)x+96+300
solving for x we get:
11/20x=396
x=720
thus the original area of corn was 720 acres
Original area of wheat was 720+480=1200 acres
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Answer:

18/5 or 36 necklace

Step-by-step explanation:

kendell made 2/5 necklace in 1/9 hours

how many will she make in 1 hour

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2/5 = 1/9

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2/5 × 1 = 1/9x

x = 2/5 ÷ 1/9

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4 0
2 years ago
39. Kate recorded the time it took six children of
scZoUnD [109]

Answer:

y=-11x+261

Step-by-step explanation:

As you can observe in the image attached, the line that best fits passes through point B and C. That means we can use those point to find the slope of such line.

m=\frac{y_{2} -y_{1} }{x_{2}-x_{1}  }

Where B(11,137) and C(12,126)

m=\frac{126-137}{12-11}=-11

So, the slope of the line that best fits is -11, approximately.

Now, we use the point-slope formula to find the equation.

y-y_{1} =m(x-x_{1} )\\y-137=-11(x-11)\\y=-11x+124+137\\y=-11x +261

Therefore, the line that best fits is y=-11x+261 approximately.

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4 0
3 years ago
X + 1 / 8 y = 2 0 x + 1 / 4 y = 4
UkoKoshka [18]

Answer:

Is that the question

Step-by-step explanation:

No seresly is that the question

4 0
3 years ago
Read 2 more answers
ANSWER PLZ QUICK AND FAST
Sergeu [11.5K]

its a. numerical expression.

4 0
2 years ago
What is the sum of a 12-term arithmetic sequence where the last term is 13 and the common difference is -10?
Y_Kistochka [10]
\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=12\\
d=-10\\
a_{12}=13
\end{cases}
\\\\\\
a_{12}=a_1+(12-1)d\implies 13=a_1+(12-1)(-10)
\\\\\\
13=a_1-110\implies \boxed{123=a_1}

\bf \\\\
-------------------------------\\\\
\textit{sum of a finite arithmetic sequence}\\\\
S_n=\cfrac{n}{2}(a_1+a_n)\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
a_n=\textit{value of the }n^{th}\ term\\
------------\\
n=12\\
a_1=123\\
a_{12}=13
\end{cases}
\\\\\\
S_{12}=\cfrac{12}{2}(a_1+a_{12})\implies S_{12}=\cfrac{12}{2}(123+13)

and surely you know how much that is.
6 0
3 years ago
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