In every case, you're finding the surface area of a rectangular prism. That area is the sum of the areas of the 6 rectangular faces. Since opposite faces have the same area, the formula can be written
... S = 2(LW +WH +HL)
The number of multiplications can be reduced if you rearrange the formula to
... S = 2(LW +H(L +W))
where L, W, and H are the length, width, and height of the prism. (It does not matter which dimension gets what name, as long as you use the same number for the same variable in the formula.)
When you're evaluating this formula over and over for diffferent sets of numbers, it is convenient to let a calculator or spreadsheet program do it for you.
1. S = 2((5 cm)(5 cm) +(5 cm)(5 cm +5 cm)) = 2(25 cm² +(5 cm)(10 cm))
... = 2(25 cm² + 50 cm²) = 150 cm²
2. S = 2(12·6 + 2(12+6)) mm² = 2(72 +36) mm² = 216 mm²
3. S = 2(11·6 + 4(11 +6)) ft² = 2·134 ft² = 264 ft²
4. S = 2(10·4 +3(10 +4)) in² = 164 in²
Answer:
depends
Step-by-step explanation:
Answer:f(x) = x becomes g(x) = (1/5)x through vertical compression by a factor of 5.
Step-by-step explanation:
Answer:
<u>Alternative hypothesis 1</u>: the mean amperage at which the fuses burn out is > 40 amperes.
<u>Alternative hypothesis 2</u>: the mean amperage at which the fuses burn out is < 40 amperes.
Step-by-step explanation:
Recall that the null hypothesis is the fact you want to refute and is in doubt.
So, in this specific case, <em>the null hypothesis would be that the mean amperage at which the fuses burn out is 40 amperes.
</em>
The alternative hypothesis are those that want to refute the null hypothesis, in this case there are 2:
<u>Alternative hypothesis 1:</u> the mean amperage at which the fuses burn out is > 40 amperes.
<u>Alternative hypothesis 2:</u> the mean amperage at which the fuses burn out is < 40 amperes.
Answer:
VA = 2/3πrB²hB
Step-by-step explanation:
Let the volume of cosA ad cone B be expressed as;
Va = 1/3πra²ha
Vb = 1/3πrb²hb
If the radius of the base of cone A is twice as large as the radius of the base of cone B
rA = 2rB
If the height of cone B is twice the height of cone A, then;
hB = 2hA
hA = hB/2
Recall that;
VA = 1/3πrA²hA
VA = 1/3π(2rB)²(hB/2)
VA = 1/3π*4rB²(hB/2)
VA = 2/3πrB²hB
This gives the volume of cone A