Answer:
no
Step-by-step explanation:
The second purchase is exactly half-again larger than the first purchase, so sheds no light on the relative costs of the items. The per-item costs can be found when the ratio of items purchased is different from one buy to another.
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We can only say that the cost of 1 pear and 3 apples is $7.
Answer:
1. x = 10
2. x = 4
Step-by-step explanation:
I use the angle ABC method:
AB² + AC² = BC²
6² + 8² = x²
x = 10
AB² + AC² = BC²
3² + x² = 5²
x = 4
<em>H</em><em>O</em><em>P</em><em>E</em><em> </em><em>T</em><em>H</em><em>I</em><em>S</em><em> </em><em>H</em><em>E</em><em>L</em><em>P</em><em>S</em><em> </em><em>A</em><em>N</em><em>D</em><em> </em><em>H</em><em>A</em><em>V</em><em>E</em><em> </em><em>A</em><em> </em><em>N</em><em>I</em><em>C</em><em>E</em><em> </em><em>D</em><em>A</em><em>Y</em><em> </em><em><</em><em>3</em>
Answer:
Step-by-step explanation:
54.08
Answer:
a = 3
Step-by-step explanation:
Since x = 7 is a solution, substitute x = 7 into the equation and solve for a
4(7) - 2(7 + a) = 8 , that is
28 - 2(7 + a) = 8 ( subtract 28 from both sides )
- 2(7 + a) = - 20 ( divide both sides by - 2 )
7 + a = 10 ( subtract 7 from both sides )
a = 3
<u>Answer:</u>
- 6. 6.75
- 7. 27
- 8. 92/17
- 9. 49
<u>Step-by-step explanation:</u>
<u>- Question 6 -</u>
- 9/16 = x/12
- => 16x = 12 x 9
- => 16x = 108
- => x = 6.75
Hence, <u>the value of x in this proportion is 6.75.</u>
<h3>_______________________________</h3>
<u>- Question 7 -</u>
- -3 + x/18 = 12/9
- => -3/18 + x/18 = 4/3
- => x/18 = 4/3 + 1/6
- => x/18 = 8/6 + 1/6
- => x/18 = 9/6
- => x = 9/6 x 18
- => x = 27
Hence, <u>the value of x is 27</u>
<h3>_______________________________</h3>
<u>- Question 8 -</u>
17/15 = 10/2x - 2
=> 17(2x - 2) = 150
=> 34x - 34 = 150
=> 34x = 184
=> x = 184/34
=> x = 92/17
Hence, <u>the value of x is 92/17</u>
<h3>_______________________________</h3>
<u>- Question 9 -</u>
- x - 16/x + 6 = 3/5
- => 5(x - 16) = 3(x + 6)
- => 5x - 80 = 3x + 18
- => 2x = 98
- => x = 49
Hence, <u>the value of x is 49.</u>
<h3>_______________________________</h3>
Hoped this helped.
