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Kitty [74]
3 years ago
10

an apple orchard contains a hundred and twelve trees the number of trees in each row is 2 less than twice the number of rows fin

d the number of rows in the number of trees per row
Mathematics
1 answer:
san4es73 [151]3 years ago
6 0
Let x be the number of rows and y be the number of trees per row.

x* y = 112
and y =  2x - 2
from first equation  y = 112/x
so  112/x = 2x - 2
112 = 2x^2 - 2x
2x^2 - 2x - 112 = 0
x^2 - x - 56 = 0
(x - 8)(x + 7) = 0
x = 8 , - 7   (ignore the negative)
so the number of rows = 8 
and number of trees per row = 112/8 = 14

Answer number of rows = 8 and number trees i each row = 14.



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Jose makes $10000 a month plus some money by commission rates. He gets 6% of everything he sells. If Jose sold $59000 worth of i
IgorC [24]

Answer:

$13540

Step-by-step explanation:

Divide 100 from $59000 in which you get 1% of that worth.  If he gets 6% of that commission rate then he would get $3540 since 1% is $590 and you multiply by 6 to get that number.  Since he makes $10000 a month as stated, you would then add the amount $3540 for his salary to get $13540.

3 0
2 years ago
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Choose the answer that validates that the rate of change is constant by showing that the ratios of the two quantities are propor
docker41 [41]

Given:

The table of values is

Number of Students   :  7    14   21    28

Number of Textbooks : 35  70  105  140

To find:

The rate of change and showing that the ratios of the two quantities are proportional and equivalent to the unit rate.

Solution:

The ratio of number of textbooks to number of students are

\dfrac{35}{7}=5

\dfrac{70}{14}=5

\dfrac{105}{21}=5

\dfrac{140}{28}=5

All the ratios of the two quantities are proportional and equivalent to the unit rate.

Let y be the number of textbooks and x be the number of students, then

\dfrac{y}{x}=k

Here, k=5.

\dfrac{y}{x}=5

y=5x

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8 0
3 years ago
Can you help me with my work
andrezito [222]

Answer:

1. x = ±9

2. x=\pm \sqrt{13}

3. 12 and -12.

4. Antoine is incorrect. There exists two solutions x=5 and x= -5.

Step-by-step explanation:

According to the questions,

Problem 1. x^{2}-81=0 i.e. x^{2}=81 i.e. x = ±9.

Problem 2. 2x^{2}-26=0 i.e. x^{2}-13=0 i.e. x^{2}=13 i.e. x=\pm \sqrt{13}

Problem 3. [tex]f(x)=x^{2}-144[tex]

To find the roots, we take, [tex]x^{2}-144=0[tex] i.e. [tex]x^{2}=144[tex] i.e. x = ±12.

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Problem 4. We have [tex]f(x)=x^{2}+25[tex]

For the roots, we take, [tex]x^{2}+25=0[tex] i.e. [tex]x^{2}=25[tex] i.e. x = ±5.

Thus, Antoine is not correct and two solutions namely x=5 and x= -5 exists.

8 0
3 years ago
Oliver uses the greatest common factor and the distributive property to rewrite this sum:
dalvyx [7]
The greatest common factor of 64 and 96 is 32, so, the first step after finding the great common factor is convert and rewrite "64 + 96" into distributive property. 

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Answer:

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Step-by-step explanation:

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