Answer:
25 gallons
Step-by-step explanation:
Given:
total no. of gallons= 50
Percentage of sulfuric acid= 50%
Number of gallons of sulfuric acid= total no. of gallons x Percentage of sulfuric acid
Putting values in above equation:
Number of gallons of sulfuric acid= 50 x 50/100
= 50 x 1/2
= 25 !
<h2>log
(6
)
+
log
(
8
)
−
log
(
2
)
</h2><h2 /><h2> Answer:</h2>
Exact Form: ㏒
(
24
)
Decimal Form: 1.38021124
<h2>
Explanation:</h2><h3> Use the product property of logarithms, </h3>
㏒b
(
x) + ㏒b
(
y
) = ㏒b
(
x
y
).
㏒
(
6
⋅
8
)
− ㏒
(
2
)
.
<h3> ⇒Use the quotient property of logarithms,</h3>
㏒
b
(
x
)
−
㏒
b
(
y
)
=
㏒
b
(
x
y
)
.
㏒
(
6
⋅
8/
2
)
<h3> ⇒Reduce the expression by cancelling the common factors.
</h3>
Factor 2 out of 6
⋅8
.
log
(
2
(
3
⋅
8
) / 2
)
<h3> Divide 3
⋅
8
by 1
.</h3>
㏒
(
3
⋅
8
)
Multiply 3 by 8
.
㏒
(
24
)
<h3>The result can be shown in both exact and decimal forms.
</h3><h3> Exact Form: ㏒
(
24
)
</h3><h3> Decimal Form: 1.38021124</h3>
Answer:
- The smallest area the field could be is 6,400 m²
- The largest area the field could be is 8,250 m²
Step-by-step explanation:
Given;
smallest possible length of the international soccer field, L₀ = 100 m
smallest possible breadth of the international soccer field, B₀ = 64 m
Largest possible length of the international soccer field, L₁ = 110 m
Largest possible breadth of the international soccer field, B₁ = 75 m
Area of a rectangle is given by;
A = L x B
The smallest area the field could be is calculated as;
A₀ = L₀ x B₀
A₀ = 100 m x 64 m
A₀ = 6,400 m²
The largest area the field could be is calculated as;
A₁ = L₁ x B₁
A₁ = 110 m x 75 m
A₁ = 8,250 m²
So we are given the system:
Written in matrix form we get:
![\left[\begin{array}{cc}2&4\\6&3\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}8\\-3\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%264%5C%5C6%263%5Cend%7Barray%7D%5Cright%5D%20%0A%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%20)
We compute the solution like this:
![ \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{cc}2&4\\6&3\end{array}\right] ^{-1} \left[\begin{array}{c}8\\-3\end{array}\right] \\= \left[\begin{array}{cc}-3&4\\6&-2\end{array}\right] \left[\begin{array}{c}8\\-3\end{array}\right] \dfrac{1}{18}\\= \left[\begin{array}{c}2\\-3\end{array}\right]](https://tex.z-dn.net/?f=%20%0A%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%0A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%264%5C%5C6%263%5Cend%7Barray%7D%5Cright%5D%20%5E%7B-1%7D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%3D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-3%264%5C%5C6%26-2%5Cend%7Barray%7D%5Cright%5D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C-3%5Cend%7Barray%7D%5Cright%5D%20%5Cdfrac%7B1%7D%7B18%7D%5C%5C%3D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C-3%5Cend%7Barray%7D%5Cright%5D)
The solution is :
![\left[\begin{array}{c}2\\-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C-3%5Cend%7Barray%7D%5Cright%5D)
