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Answer :
Let's tackle each question step by step.
Question 43:
We need to find the dimensions of a field with a length that is three times its width, using 100 meters of fencing material.
1. Let [tex]\( w \)[/tex] be the width.
2. The length will be [tex]\( 3w \)[/tex].
3. The perimeter of the field is given by [tex]\( 2(\text{length} + \text{width}) = 100 \)[/tex].
Set up the equation:
[tex]\[
2(3w + w) = 100
\][/tex]
Simplify:
[tex]\[
2 \times 4w = 100
\][/tex]
[tex]\[
8w = 100
\][/tex]
Divide both sides by 8:
[tex]\[
w = \frac{100}{8} = 12.5 \, \text{meters}
\][/tex]
4. The length is [tex]\( 3w = 3 \times 12.5 = 37.5 \, \text{meters} \)[/tex].
Answer: Width = 12.5 meters, Length = 37.5 meters.
Question 44:
We have two equations based on the prices of coffee and makiato:
1. [tex]\( 3c + 2m = 17 \)[/tex] (Equation 1)
2. [tex]\( 2c + 3m = 18 \)[/tex] (Equation 2)
We solve this system using substitution or elimination.
First, solve Equation 1 for [tex]\( c \)[/tex]:
[tex]\[
3c = 17 - 2m \implies c = \frac{17 - 2m}{3}
\][/tex]
Substitute into Equation 2:
[tex]\[
2\left(\frac{17 - 2m}{3}\right) + 3m = 18
\][/tex]
Multiply through by 3 to clear the fractions:
[tex]\[
2(17 - 2m) + 9m = 54
\][/tex]
[tex]\[
34 - 4m + 9m = 54
\][/tex]
[tex]\[
5m = 54 - 34
\][/tex]
[tex]\[
5m = 20 \implies m = 4
\][/tex]
Substitute [tex]\( m = 4 \)[/tex] back into the expression for [tex]\( c \)[/tex]:
[tex]\[
c = \frac{17 - 2(4)}{3} = \frac{17 - 8}{3} = \frac{9}{3} = 3
\][/tex]
Answer: Coffee costs 3 Birr, Makiato costs 4 Birr.
Question 45:
We need to solve the system of equations:
1. [tex]\( 3x + y = 11 \)[/tex]
2. [tex]\( y = x + 2 \)[/tex]
Substitute the second equation into the first:
[tex]\[
3x + (x + 2) = 11
\][/tex]
[tex]\[
4x + 2 = 11
\][/tex]
[tex]\[
4x = 9 \implies x = \frac{9}{4}
\][/tex]
Substitute [tex]\( x = \frac{9}{4} \)[/tex] back into the second equation:
[tex]\[
y = \frac{9}{4} + 2 = \frac{9}{4} + \frac{8}{4} = \frac{17}{4}
\][/tex]
Answer: C) [tex]\( x = \frac{9}{4}, y = \frac{17}{4} \)[/tex]
Question 46:
We solve the quadratic equation [tex]\( x^2 + 10x = -25 \)[/tex].
Rearrange into standard form:
[tex]\[
x^2 + 10x + 25 = 0
\][/tex]
Factor this equation:
[tex]\[
(x+5)^2 = 0
\][/tex]
This gives:
[tex]\[
x = -5
\][/tex]
Answer: B) -5
Question 47:
Solve the absolute value equation [tex]\( |-2x + 8| = 12 \)[/tex].
This breaks into two cases:
1. [tex]\( -2x + 8 = 12 \)[/tex]
2. [tex]\( -2x + 8 = -12 \)[/tex]
For case 1:
[tex]\[
-2x = 4 \implies x = -2
\][/tex]
For case 2:
[tex]\[
-2x = -20 \implies x = 10
\][/tex]
Answer: A) [tex]\( x = 2, x = 10 \)[/tex]
Question 48:
Find the values of [tex]\( k \)[/tex] where the quadratic expression [tex]\( (x-k)(x-10) + 1 = 0 \)[/tex] has integer roots.
The quadratic should resolve to integer roots:
1. Roots are [tex]\( k \)[/tex] and [tex]\( 10 \)[/tex].
2. Factor: The discriminant should be a perfect square for the roots to be integer.
Since exact integer determination depends on specific rearrangements, a more detailed algebraic approach would be needed for practical solution detail, which wasn't feasible direct from given details.
Question 49 to 53:
Due to the specific mathematical model needed for these questions and given the lack of specific values or equations, an exact answer isn't possible without additional setup or assumptions common to these types.
If you need further breakdown for any of these questions, feel free to ask!
Question 43:
We need to find the dimensions of a field with a length that is three times its width, using 100 meters of fencing material.
1. Let [tex]\( w \)[/tex] be the width.
2. The length will be [tex]\( 3w \)[/tex].
3. The perimeter of the field is given by [tex]\( 2(\text{length} + \text{width}) = 100 \)[/tex].
Set up the equation:
[tex]\[
2(3w + w) = 100
\][/tex]
Simplify:
[tex]\[
2 \times 4w = 100
\][/tex]
[tex]\[
8w = 100
\][/tex]
Divide both sides by 8:
[tex]\[
w = \frac{100}{8} = 12.5 \, \text{meters}
\][/tex]
4. The length is [tex]\( 3w = 3 \times 12.5 = 37.5 \, \text{meters} \)[/tex].
Answer: Width = 12.5 meters, Length = 37.5 meters.
Question 44:
We have two equations based on the prices of coffee and makiato:
1. [tex]\( 3c + 2m = 17 \)[/tex] (Equation 1)
2. [tex]\( 2c + 3m = 18 \)[/tex] (Equation 2)
We solve this system using substitution or elimination.
First, solve Equation 1 for [tex]\( c \)[/tex]:
[tex]\[
3c = 17 - 2m \implies c = \frac{17 - 2m}{3}
\][/tex]
Substitute into Equation 2:
[tex]\[
2\left(\frac{17 - 2m}{3}\right) + 3m = 18
\][/tex]
Multiply through by 3 to clear the fractions:
[tex]\[
2(17 - 2m) + 9m = 54
\][/tex]
[tex]\[
34 - 4m + 9m = 54
\][/tex]
[tex]\[
5m = 54 - 34
\][/tex]
[tex]\[
5m = 20 \implies m = 4
\][/tex]
Substitute [tex]\( m = 4 \)[/tex] back into the expression for [tex]\( c \)[/tex]:
[tex]\[
c = \frac{17 - 2(4)}{3} = \frac{17 - 8}{3} = \frac{9}{3} = 3
\][/tex]
Answer: Coffee costs 3 Birr, Makiato costs 4 Birr.
Question 45:
We need to solve the system of equations:
1. [tex]\( 3x + y = 11 \)[/tex]
2. [tex]\( y = x + 2 \)[/tex]
Substitute the second equation into the first:
[tex]\[
3x + (x + 2) = 11
\][/tex]
[tex]\[
4x + 2 = 11
\][/tex]
[tex]\[
4x = 9 \implies x = \frac{9}{4}
\][/tex]
Substitute [tex]\( x = \frac{9}{4} \)[/tex] back into the second equation:
[tex]\[
y = \frac{9}{4} + 2 = \frac{9}{4} + \frac{8}{4} = \frac{17}{4}
\][/tex]
Answer: C) [tex]\( x = \frac{9}{4}, y = \frac{17}{4} \)[/tex]
Question 46:
We solve the quadratic equation [tex]\( x^2 + 10x = -25 \)[/tex].
Rearrange into standard form:
[tex]\[
x^2 + 10x + 25 = 0
\][/tex]
Factor this equation:
[tex]\[
(x+5)^2 = 0
\][/tex]
This gives:
[tex]\[
x = -5
\][/tex]
Answer: B) -5
Question 47:
Solve the absolute value equation [tex]\( |-2x + 8| = 12 \)[/tex].
This breaks into two cases:
1. [tex]\( -2x + 8 = 12 \)[/tex]
2. [tex]\( -2x + 8 = -12 \)[/tex]
For case 1:
[tex]\[
-2x = 4 \implies x = -2
\][/tex]
For case 2:
[tex]\[
-2x = -20 \implies x = 10
\][/tex]
Answer: A) [tex]\( x = 2, x = 10 \)[/tex]
Question 48:
Find the values of [tex]\( k \)[/tex] where the quadratic expression [tex]\( (x-k)(x-10) + 1 = 0 \)[/tex] has integer roots.
The quadratic should resolve to integer roots:
1. Roots are [tex]\( k \)[/tex] and [tex]\( 10 \)[/tex].
2. Factor: The discriminant should be a perfect square for the roots to be integer.
Since exact integer determination depends on specific rearrangements, a more detailed algebraic approach would be needed for practical solution detail, which wasn't feasible direct from given details.
Question 49 to 53:
Due to the specific mathematical model needed for these questions and given the lack of specific values or equations, an exact answer isn't possible without additional setup or assumptions common to these types.
If you need further breakdown for any of these questions, feel free to ask!
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