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7. How many ounces of tin and copper, each, in 56 ounces of gun metal, if gun metal consists of 1 part of tin and 9 parts of copper? 8. The sides of a triangle are 8, 12, and 16 inches; the shortest side of a similar triangle is 10 inches. Find the remaining sides of the similar triangle.

NOTE. In two similar triangles any two sides have the same ratio as the corresponding sides of the other.

9. The diameter of the standard baseball is 3 inches. How does its surface compare with the surface of a slate globe 2 ft. in diameter?

NOTE. The areas of similar figures are to each other as the squares on corresponding lines.

10. What proportion exists between the times t1 and t2 it will take to do a certain piece of work if the number of men employed in the first case is n1 and in the second case is n2?

11. What proportion exists between the numbers of sheep that can be grazed in two square fields, one having a side of length a and the other a side of length b? Use n1 and n to represent the two numbers.

SUPPLEMENTARY EXERCISES ON §§ 123–124

Draw the graph of each of the following equations.

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Solve each of the following simultaneous equations by first draw

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SUPPLEMENTARY EXERCISES ON §§ 128-130

Solve (by any method) the following simultaneous equations.

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15. A workman is engaged for 15 days; he receives $2 for every day he works and agrees to forfeit 50 cents a day for every day he is idle. At the end of the time he receives $22.50. How many days

did he work?

16. Two boys, A and B, wishing to determine their weights, find that they balance on a teeter board when B is 6 feet from the fulcrum and A is 5 feet from it. If B carries a 30-pound weight with him, they balance when B is 4 feet and A 5 feet from the fulcrum. How heavy is each boy?

SUPPLEMENTARY EXERCISES ON § 136

Reduce to simpler form each of the following expressions.

1. √64. 2. Vasc. 3. √900 r2s6. 4. Van 5. √25 a6b4c10

Find the square root of each of the following expressions.

6. 25 x2-80 xy+64 y2.

7. 16 x2-2 xy+1% y2.

8. 4 x2+9 y2+16–12 xy+16 x−24 y. ·

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Solve each of the following equations. Check each answer.

1. Vx2+9=5. 2. √x2-36-8=0.

3. √2x-2=√2x+10-2.

6. 3x-√9 x2 – 12 x−51=3.

7. x−1−√x2+x-11=0.

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Rationalize the denominator in each of the following expressions.

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Solve each of the following equations, finding the two roots.

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x+25 1+x

7. 22+2=5.

10. ax2+c=b.

SUPPLEMENTARY EXERCISES ON § 155

Solve each of the following quadratics by completing the square.

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SUPPLEMENTARY EXERCISES ON §§ 156-157

Solve each of the following equations.

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11. If a wire l feet long is stretched between two poles of equal height which are H feet apart, the sag d, which the wire has at its middle point is given by the formula

d = √
3 HI-3 H2
8

ft.

By use of this formula find the actual length of wire needed to stretch between two poles which are 150 feet apart provided a sag of 5 feet is to be allowed. 1=150.44+ ft. Ans.

12. The figure represents an elevated water tank such as is commonly used at manufacturing plants. The lower part consists

FIG. 96.

of a hemispherical bowl upon which rests a
cylindrical part and at the top is a conical roof.
The water occupies the bowl and the cylinder,
but not the cone. If the height of the cylinder
be 1 times its diameter, what must be the
diameter in order that the tank may hold
24,200 gallons? (Work in feet, allowing .14
cubic foot to a gallon and use the tables for
the extraction of necessary cube roots.)
2294 13.2988 ft. Ans.

13. The sum of the areas of two circles is A and the sum of their radii is s. Find the formulas giving the radius of each.

PART II. TABLE OF POWERS AND ROOTS

EXPLANATION

1. Square Roots. The way to find square roots from the Table is best understood from an example. Thus, suppose we wish to find V1.48. To do this we first locate 1.48 in the column headed by the letter n. We find it near the bottom of this column (next to the last number). Now we go across on that level until we get into the column headed by Vn. We find at that place the number 1.21655. This is our answer. That is, V1.48=1.21655 (approximately).

If we had wanted V14.8 instead of V1.48 the work would have been the same except that we would have gone over into the column headed 10 n (because 14.8=10×1.48). The number thus located is seen to be 3.84708, which is, therefore, the desired value of √14.8.

Again, if we had wished to find V148 the work would take us back again to the column headed √n, but now instead of the answer being 1.21655 it would be 12.1655. In other words, the order of the digits in √148 is the same as for √1.48, but the decimal point in the answer is one place farther to the right.

Similarly, if we desired

1480 the work would be the same as before except that we must now use the column headed ✓10 n and move the decimal point there occurring one place farther to the right. This is seen to give 38.4708.

Thus we see how to get the square root of 1.48 or any power of 10 times that number.

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