6(a+3)=18+6a soolve for a
2 answers:
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To solve, simply do this:
Then you'll get the answer, -7/16
The volume of a cuboid is given by length × width × height
We have:
Volume = 7.6 ft³
Height = 3x - 1
Length = x + 5
Width = x
Substituting these into the formula, we have:
7.6 = (3x - 1) (x + 5) (x)
7.6 = [3x² + 15x - x - 5] (x)
7.6 = [3x² + 14x - 5](x)
7.6 = 3x³ + 14x² - 5x
0 = 3x³ + 14x² - 5x - 7.6
Drawing the graph is one way of finding the solution (refer to the graph below):
We have three solutions (where the curve crosses the x-axis):
x = -4.9
x = -0.6
x = 0.8
Putting these solutions back into the context, since we are looking for the value of x which is part of measurement of length, we cannot have negative value, so we will take the value of x = 0.8 ft
Converting 0.8 ft into inches = 0.8 × 12 inches = 9.6 inches
Answer: x = 9.6 inches
We have:
Volume = 7.6 ft³
Height = 3x - 1
Length = x + 5
Width = x
Substituting these into the formula, we have:
7.6 = (3x - 1) (x + 5) (x)
7.6 = [3x² + 15x - x - 5] (x)
7.6 = [3x² + 14x - 5](x)
7.6 = 3x³ + 14x² - 5x
0 = 3x³ + 14x² - 5x - 7.6
Drawing the graph is one way of finding the solution (refer to the graph below):
We have three solutions (where the curve crosses the x-axis):
x = -4.9
x = -0.6
x = 0.8
Putting these solutions back into the context, since we are looking for the value of x which is part of measurement of length, we cannot have negative value, so we will take the value of x = 0.8 ft
Converting 0.8 ft into inches = 0.8 × 12 inches = 9.6 inches
Answer: x = 9.6 inches
Number of dimes were 81 and number of quarters were 47
<em><u>Solution:</u></em>
Let "d" be the number of dimes
Let "q" be the number of quarters
We know that,
value of 1 dime = $ 0.10
value of 1 quarter = $ 0.25
<em><u>Given that There are 128 coins in all</u></em>
number of dimes + number of quarters = 128
d + q = 128 ------ eqn 1
<em><u>Also given that collection of dimes and quarters is worth $19.85</u></em>
number of dimes x value of 1 dime + number of quarters x value of 1 quarter = 19.85
0.1d + 0.25q = 19.85 -------- eqn 2
<em><u>Let us solve eqn 1 and eqn 2</u></em>
From eqn 1,
d = 128 - q -------- eqn 3
<em><u>Substitute eqn 3 in eqn 2</u></em>
0.1(128 - q) + 0.25q = 19.85
12.8 - 0.1q + 0.25q = 19.85
12.8 + 0.15q = 19.85
0.15q = 7.05
<h3>q = 47</h3>Therefore from eqn 3,
d = 128 - q
d = 128 - 47
<h3>d = 81</h3>Thus number of dimes were 81 and number of quarters were 47