Answer:
1
Step-by-step explanation:
6/2(1+2)
6/2+4
6/6
1
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Answer:
In inequality notation:
Domain: -1 ≤ x ≤ 3
Range: -4 ≤ x ≤ 0
In set-builder notation:
Domain: {x | -1 ≤ x ≤ 3 }
Range: {y | -4 ≤ x ≤ 0 }
In interval notation:
Domain: [-1, 3]
Range: [-4, 0]
Step-by-step explanation:
The domain is all the x-values of a relation.
The range is all the y-values of a relation.
In this example, we have an equation of a circle.
To find the domain of a relation, think about all the x-values the relation can be. In this example, the x-values of the relation start at the -1 line and end at the 3 line. The same can be said for the range, for the y-values of the relation start at the -4 line and end at the 0 line.
But what should our notation be? There are three ways to notate domain and range.
Inequality notation is the first notation you learn when dealing with problems like these. You would use an inequality to describe the values of x and y.
In inequality notation:
Domain: -1 ≤ x ≤ 3
Range: -4 ≤ x ≤ 0
Set-builder notation is VERY similar to inequality notation except for the fact that it has brackets and the variable in question.
In set-builder notation:
Domain: {x | -1 ≤ x ≤ 3 }
Range: {y | -4 ≤ x ≤ 0 }
Interval notation is another way of identifying domain and range. It is the idea of using the number lines of the inequalities of the domain and range, just in algebriac form. Note that [ and ] represent ≤ and ≥, while ( and ) represent < and >.
In interval notation:
Domain: [-1, 3]
Range: [-4, 0]
Answer:
2i: 169.71
2ii: 0.17L
3a: 4×10⁻⁵
3b: 110011
Step-by-step explanation:
2i. The surface of the top and bottom of the tin is two times (top and bottom) π·r² = 2·π·3² = 18π cm².
The circumference of the circle is 2·π·r = 6π cm².
The area of the material connecting top and bottom is a rectangle of the tin height times the circumference: 6·6π = 36π cm².
This gives a total of 18π + 36π = 54π cm².
With π approximated by 22/7 the total surface area is 54*22/7 ≈ 169.71.
Notice how the calculation is simple by waiting until the very last moment to substitute π.
2ii. The volume is the area π·r² of the circle times the height of the tin: 9π*6 = 54π cm³ ≈ 169.71 cm³.
Since 1L = 1000 cm³ the volume is 0.16971 litres, which should be rounded to 0.17 L.
3a: If we rewrite P as 36 x 10⁻⁴ and realize that 36/2.25 = 16, then the fraction can be written as
16 x 10⁻⁴⁻⁶ = 16 x 10⁻¹⁰.
The square root of that is taking it to the power of 1/2, so (16x10⁻¹⁰)^0.5 = 4x10⁻⁵ = 0.00004
3b: 1111 1111 is 255 in decimal. 101 is 5 in decimal. 255/5 is 51 in decimal. 51 in binary is 110011.
Answer:
draw the diagram and multiply 5x20
