Tüm alıştırma soruları

541 soru

Soru 441Soru

A water purification plant processes raw water through three successive filtration stages: Stage A, Stage B, and Stage C. Stage A removes 38\frac{3}{8} of the impurities present in the raw water. Stage B removes 40%40\% of the remaining impurities. Stage C removes 0.750.75 of the impurities that remain after Stage B. If 4.54.5 kilograms of impurities are successfully removed in Stage C, how many kilograms of impurities were originally in the raw water?

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Cevap: 16

Cevap

The original amount of impurities in the raw water was 1616 kilograms.
The correct answer is 1616 kg. To find this, we express the remaining impurities at each stage as a fraction of the initial amount xx. After Stage A, 58x\frac{5}{8}x remains. In Stage B, 40%40\% is removed, meaning 60%60\% of 58x\frac{5}{8}x, or 38x\frac{3}{8}x, remains. In Stage C, 0.750.75 (or 34\frac{3}{4}) of this remainder is removed, which is 34×38x=932x\frac{3}{4} \times \frac{3}{8}x = \frac{9}{32}x. Setting 932x=4.5\frac{9}{32}x = 4.5 gives x=16x = 16.

Adım Adım Çözüm

1
Represent the initial mass of impurities in the raw water with a variable.
Let xx be the initial kilograms of impurities.
Establishing a variable allows setting up an equation to track the changes through each stage.
2
Calculate the remaining impurities after Stage A.
Impurities remaining = x38x=58xx - \frac{3}{8}x = \frac{5}{8}x kg.
Stage A removes 38\frac{3}{8} of the initial impurities, so 138=581 - \frac{3}{8} = \frac{5}{8} of the impurities remain.
3
Calculate the impurities removed and remaining after Stage B.
Impurities removed in Stage B = 0.40×58x=14x0.40 \times \frac{5}{8}x = \frac{1}{4}x kg. Impurities remaining after Stage B = \frac{5}{8}x - \frac{1}{4}x = \frac{3}{8}x$ kg.
Stage B removes 40%40\% of the impurities that remained after Stage A. Subtracting the removed portion from the starting amount for this stage yields the remaining fraction.
4
Express the impurities removed in Stage C.
Impurities removed in Stage C = 0.75×38x=932x0.75 \times \frac{3}{8}x = \frac{9}{32}x kg.
Stage C removes 0.750.75 (or 34\frac{3}{4}) of the impurities remaining after Stage B.
5
Equate the Stage C expression to the given value and solve for xx.
932x=4.5    x=4.5×329=16\frac{9}{32}x = 4.5 \implies x = 4.5 \times \frac{32}{9} = 16 kg.
The problem states that 4.54.5 kg of impurities are removed in Stage C, so solving this equation yields the initial value.

Anahtar Kavram

Solving multi-step word problems involving successive applications of fractions, decimals, and percentages.

Alternatif Yöntem

Working backwards from the final stage can simplify the calculations. Since Stage C removes 0.750.75 of the impurities remaining after Stage B, and this amount equals 4.54.5 kg, the amount remaining after Stage B is 4.50.75=6\frac{4.5}{0.75} = 6 kg. Since Stage B removes 40%40\% of the impurities remaining after Stage A, the 66 kg represents 100%40%=60%100\% - 40\% = 60\% of the impurities remaining after Stage A. Thus, the amount remaining after Stage A is 60.60=10\frac{6}{0.60} = 10 kg. Finally, since Stage A removes 38\frac{3}{8} of the original impurities, the 1010 kg represents 138=581 - \frac{3}{8} = \frac{5}{8} of the original impurities. The original amount is therefore 10×85=1610 \times \frac{8}{5} = 16 kg.
Tahmini Süre:3m 0s
Soru 442Soru

Two integers, xx and yy, are positioned on a number line. The distance between xx and 11 is twice the distance between yy and 22. If the distance between xx and yy is exactly 55 units, what is the sum of all possible values of x+yx + y?

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Cevap: 17

Cevap

The sum of all possible values of x+yx + y is 1717.
Representing the distances algebraically yields the system x1=2y2|x - 1| = 2|y - 2| and xy=5|x - y| = 5. Since xx and yy must be integers, splitting these equations into positive and negative cases yields exactly three valid integer coordinate pairs: (13,8)(13, 8), (5,0)(5, 0), and (7,2)(-7, -2). The sums (x+yx + y) for these pairs are 2121, 55, and 9-9, respectively. Adding these possible sums together gives a final total of 1717.

Adım Adım Çözüm

1
Set up the absolute value expressions representing the distances.
x1=2y2|x - 1| = 2|y - 2| and xy=5|x - y| = 5
Distance on a number line between two points aa and bb is mathematically defined as ab|a - b|.
2
Split the distance condition xy=5|x - y| = 5 into two coordinate cases.
x=y+5x = y + 5 or x=y5x = y - 5
An absolute value equation of the form A=B|A| = B splits into A=BA = B or A=BA = -B.
3
Substitute x=y+5x = y + 5 into the first equation and solve for integer values of yy.
y=8y = 8 (which gives x=13x = 13) and y=0y = 0 (which gives x=5x = 5)
This generates the first set of valid integer coordinates satisfying all constraints.
4
Substitute x=y5x = y - 5 into the first equation and solve for integer values of yy.
y=2y = -2 (which gives x=7x = -7); the second algebraic option y=10/3y = 10/3 is discarded because it is not an integer
This generates the remaining valid integer coordinates satisfying all constraints.
5
Sum the value of x+yx + y for all three valid coordinate pairs.
(13+8)+(5+0)+(72)=21+59=17(13 + 8) + (5 + 0) + (-7 - 2) = 21 + 5 - 9 = 17
The question asks for the sum of all possible values of the expression x+yx + y.

Anahtar Kavram

Using absolute value to represent distances on a number line and solving systems of absolute value equations under integer constraints.
Tahmini Süre:2m 30s
Soru 443Soru

What is the sum of all integer values of yy that satisfy the inequality 2y37|2y - 3| \le 7?

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Cevap: 12

Cevap

The sum of all integer values of yy that satisfy the inequality is 12.
Solving the inequality 2y37|2y - 3| \le 7 requires setting up the compound inequality 72y37-7 \le 2y - 3 \le 7. Adding 3 to all parts gives 42y10-4 \le 2y \le 10, and dividing by 2 yields the interval 2y5-2 \le y \le 5. The integers in this closed interval are 2,1,0,1,2,3,4-2, -1, 0, 1, 2, 3, 4, and 55. Summing these values gives 12, as the terms 2-2 and 1-1 cancel out with 22 and 11.

Adım Adım Çözüm

1
Set up the compound inequality
72y37-7 \le 2y - 3 \le 7
An absolute value inequality of the form ab|a| \le b translates to bab-b \le a \le b.
2
Isolate the term containing yy
42y10-4 \le 2y \le 10
Add 3 to all three parts of the compound inequality to eliminate the 3-3.
3
Solve for yy
2y5-2 \le y \le 5
Divide all three parts of the inequality by 2.
4
Identify the integer solutions in the interval
2,1,0,1,2,3,4,5-2, -1, 0, 1, 2, 3, 4, 5
The inequality includes the endpoints, so the integers satisfying the inequality are all integers from 2-2 through 55, inclusive.
5
Calculate the sum of the integers
12
Adding the integers: (2)+(1)+0+1+2+3+4+5=12(-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 = 12. The negative integers cancel out their corresponding positive counterparts (2-2 and 22, 1-1 and 11).

Anahtar Kavram

Solving compound inequalities derived from absolute value inequalities and finding the sum of the integer solution set.
Tahmini Süre:1m 30s
Soru 444Soru

A school band has between 100100 and 150150 members. When the band members line up in rows of 66, there are 22 members left over. When they line up in rows of 99, there are also 22 members left over. When they line up in rows of 55, there are no members left over. How many members are in the school band?

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Cevap: 110

Cevap

There are 110 members in the school band.
The number of members minus 2 must be a multiple of the least common multiple of 6 and 9, which is 18. The possible values between 100 and 150 are 110, 128, and 146. Among these, only 110 is divisible by 5, which satisfies the condition of having no members left over when grouped in rows of 5.

Adım Adım Çözüm

1
Find the least common multiple (LCM) of the row sizes 6 and 9.
The LCM of 6 and 9 is 18.
Since the remainder is the same (2) for both row sizes, the number of members minus 2 must be a common multiple of 6 and 9.
2
Identify candidate numbers between 100 and 150 that are 2 more than a multiple of 18.
The candidate numbers are 110, 128, and 146.
To satisfy the condition of having a remainder of 2 when divided by 6 and 9, the total must be of the form 18k+218k + 2 within the given range [100,150][100, 150].
3
Check which candidate is divisible by 5.
110 is divisible by 5.
Since there are no members left over when lined up in rows of 5, the total number of members must be a multiple of 5.

Anahtar Kavram

Using the least common multiple (LCM) and remainders to solve divisibility word problems.
Tahmini Süre:1m 30s
Soru 445Soru

In a school auditorium, 38\frac{3}{8} of the total seats are in the balcony, and the remaining seats are on the main floor. For the spring play, 80%80\% of the balcony seats were occupied, and 60%60\% of the main floor seats were occupied. If there were exactly 117 empty seats in the auditorium, what was the total number of seats in the auditorium?

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Cevap: 360

Cevap

The total number of seats in the auditorium is 360.
The correct answer of 360 is found by expressing the empty seats from each section as a fraction of the total seats. Since the balcony represents 38\frac{3}{8} of the total seats and is 20%20\% empty, the empty balcony seats equal 340\frac{3}{40} of the total seats. The main floor represents 58\frac{5}{8} of the total seats and is 40%40\% empty, so the empty main floor seats equal 1040\frac{10}{40} of the total seats. Summing these yields 1340\frac{13}{40} of the total seats, which equals 117. Solving 1340S=117\frac{13}{40}S = 117 gives a total of 360 seats.

Adım Adım Çözüm

1
Define the variable for the total seats and find the fraction of seats in each section.
Let the total number of seats be SS. The balcony has 38S\frac{3}{8}S seats, and the main floor has 138=58S1 - \frac{3}{8} = \frac{5}{8}S seats.
Establishing fractional partitions of the total capacity is necessary to set up the equation.
2
Calculate the fraction of total seats that are empty in the balcony.
Empty balcony seats = 20%20\% of 38S=0.20×38S=15×38S=340S\frac{3}{8}S = 0.20 \times \frac{3}{8}S = \frac{1}{5} \times \frac{3}{8}S = \frac{3}{40}S.
Since 80% of the balcony seats are occupied, the remaining 20% are empty.
3
Calculate the fraction of total seats that are empty on the main floor.
Empty main floor seats = 40%40\% of 58S=0.40×58S=25×58S=1040S\frac{5}{8}S = 0.40 \times \frac{5}{8}S = \frac{2}{5} \times \frac{5}{8}S = \frac{10}{40}S.
Since 60% of the main floor seats are occupied, the remaining 40% are empty.
4
Sum the empty seat fractions and set the sum equal to the total number of empty seats.
Total empty seats = 340S+1040S=1340S\frac{3}{40}S + \frac{10}{40}S = \frac{13}{40}S. The equation is 1340S=117\frac{13}{40}S = 117.
Combining the empty seats from both sections represents the total empty seat count.
5
Solve the equation for the total number of seats SS.
S=117×4013=9×40=360S = 117 \times \frac{40}{13} = 9 \times 40 = 360.
Multiplying by the reciprocal of the empty seats fraction isolates the variable for the total seats.

Anahtar Kavram

Solving multi-step word problems involving fractions, decimal conversions, and percentages.
Soru 446Soru

On a standard number line, point AA is located at 24-24 and point BB is located at 66. Point CC is located between point AA and point BB such that the distance from AA to CC is 23\frac{2}{3} of the distance from CC to BB. What is the coordinate of point CC?

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Cevap: -12

Cevap

The coordinate of point CC is 12-12.
Because point CC with coordinate cc lies between 24-24 and 66, the distance from AA to CC is c(24)=c+24c - (-24) = c + 24, and the distance from CC to BB is 6c6 - c. Setting the distance from AA to CC to 23\frac{2}{3} of the distance from CC to BB gives the equation c+24=23(6c)c + 24 = \frac{2}{3}(6 - c). Multiplying both sides by 33 to clear the fraction results in 3c+72=122c3c + 72 = 12 - 2c. Collecting like terms yields 5c=605c = -60, which simplifies to c=12c = -12.

Adım Adım Çözüm

1
Express the distances between the points on the number line using their coordinates.
The distance from AA to CC is c+24c + 24, and the distance from CC to BB is 6c6 - c.
Since point CC is positioned between points AA and BB, the inequality 24<c<6-24 < c < 6 holds. This allows the absolute value distance expressions c(24)|c - (-24)| and 6c|6 - c| to simplify directly to positive expressions without absolute value bars.
2
Formulate an equation based on the specified ratio of distances.
c+24=23(6c)c + 24 = \frac{2}{3}(6 - c)
The problem states that the distance from AA to CC is 23\frac{2}{3} of the distance from CC to BB.
3
Solve the linear equation for the coordinate cc.
c=12c = -12
Multiplying both sides by 33 gives 3(c+24)=2(6c)3(c + 24) = 2(6 - c), which expands to 3c+72=122c3c + 72 = 12 - 2c. Rearranging the terms by adding 2c2c to both sides and subtracting 7272 from both sides results in 5c=605c = -60. Dividing by 55 gives c=12c = -12.

Anahtar Kavram

Using absolute value properties to express distances on a number line and solving partitioning coordinate problems.
Tahmini Süre:1m 15s
Soru 447Soru

A set of cards is numbered consecutively from 11 to NN. If exactly 1212 of these cards have a number that is a multiple of both 66 and 88, what is the greatest possible value of NN?

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Cevap: 311

Cevap

The greatest possible value of NN is 311311.
The correct value is 311. Since the numbers must be multiples of both 6 and 8, they must be multiples of their least common multiple, which is 24. The twelfth multiple of 24 is 288, and the thirteenth multiple is 312. To have exactly 12 such multiples, the maximum number N must be at least 288 but strictly less than 312, meaning the largest integer value is 311.

Adım Adım Çözüm

1
Find the least common multiple (LCM) of 6 and 8.
The LCM of 6 and 8 is 24.
Any number that is a multiple of both 6 and 8 must be a multiple of their least common multiple.
2
Calculate the 12th and 13th multiples of 24.
The 12th multiple is 12×24=28812 \times 24 = 288 and the 13th multiple is 13×24=31213 \times 24 = 312.
To have exactly 12 multiples in the set, the set must include the 12th multiple but exclude the 13th multiple.
3
Determine the maximum value of NN such that 312 is not included.
N=311N = 311.
The largest integer less than 312 is 311. If NN were 312 or greater, the set would contain 13 or more multiples.

Anahtar Kavram

Least Common Multiple and Divisibility Properties
Soru 448Soru

On a standard number line, point AA has coordinate 15-15 and point BB has coordinate 1717. Point CC is located to the right of point BB such that the distance between AA and CC is exactly 33 times the distance between BB and CC. What is the coordinate of point CC?

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Cevap: 33

Cevap

The coordinate of point CC is 33.
The correct coordinate is found by setting up the distance equation for point CC (with coordinate c>17c > 17) relative to A(15)A(-15) and B(17)B(17). The distance ACAC is c(15)=c+15c - (-15) = c + 15, and the distance BCBC is c17c - 17. Setting c+15=3(c17)c + 15 = 3(c - 17) and solving yields c=33c = 33, which is to the right of BB.

Adım Adım Çözüm

1
Define the variable for the coordinate of point C and write the expressions for distances.
Let cc be the coordinate of point CC. The distance between AA and CC is c(15)=c+15|c - (-15)| = |c + 15|, and the distance between BB and CC is c17|c - 17|. Since point CC is to the right of point BB (which is at 1717), we know c>17c > 17, so c+15=c+15|c + 15| = c + 15 and c17=c17|c - 17| = c - 17.
To set up an algebraic equation representing the physical distance relations on the number line.
2
Set up the equation using the given relationship.
The equation is c+15=3(c17)c + 15 = 3(c - 17).
The problem states the distance between AA and CC is 33 times the distance between BB and CC.
3
Solve the equation for cc.
c+15=3c51    15+51=3cc    66=2c    c=33c + 15 = 3c - 51 \implies 15 + 51 = 3c - c \implies 66 = 2c \implies c = 33.
To find the coordinate of point CC.

Anahtar Kavram

Calculating distances between points on a number line using absolute value and solving the resulting equations.

Alternatif Yöntem

Use geometric visualization: The distance from A(15)A(-15) to B(17)B(17) is 17(15)=3217 - (-15) = 32 units. Since point CC lies to the right of BB, the distance ACAC is the sum of ABAB and BCBC. Therefore, AC=32+BCAC = 32 + BC. We are given that AC=3×BCAC = 3 \times BC. Substituting this gives 32+BC=3×BC    2×BC=32    BC=1632 + BC = 3 \times BC \implies 2 \times BC = 32 \implies BC = 16. Since CC is 16 units to the right of B(17)B(17), its coordinate is 17+16=3317 + 16 = 33.
Tahmini Süre:1m 30s
Soru 449Soru

If xx is a real number such that 4x+2+4x+2+4x+2+4x+28x3=64\sqrt{\frac{4^{x+2} + 4^{x+2} + 4^{x+2} + 4^{x+2}}{8^{x-3}}} = 64, what is the value of xx?

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Cevap: 3

Cevap

The value of xx is 33.
By rewriting the repeated addition in the numerator as 4×4x+2=4x+34 \times 4^{x+2} = 4^{x+3} and converting all bases to 2, the expression inside the square root simplifies to 215x2^{15-x}. Taking the square root gives 215x22^{\frac{15-x}{2}}. Setting this equal to 6464 (which is 262^6) and equating the exponents yields x=3x = 3.

Adım Adım Çözüm

1
Rewrite the sum in the numerator 4x+2+4x+2+4x+2+4x+24^{x+2} + 4^{x+2} + 4^{x+2} + 4^{x+2} as a product.
4×4x+2=4x+34 \times 4^{x+2} = 4^{x+3}
Adding a term to itself four times is equivalent to multiplying that term by 4.
2
Convert the base 4 numerator and base 8 denominator to base 2.
Numerator: (22)x+3=22x+6(2^2)^{x+3} = 2^{2x+6}; Denominator: (23)x3=23x9(2^3)^{x-3} = 2^{3x-9}
Expressing terms with the same base allows the use of exponent laws to simplify the fraction.
3
Simplify the fraction by subtracting the denominator's exponent from the numerator's exponent.
22x+623x9=2(2x+6)(3x9)=215x\frac{2^{2x+6}}{2^{3x-9}} = 2^{(2x+6)-(3x-9)} = 2^{15-x}
The quotient rule for exponents states that aman=amn\frac{a^m}{a^n} = a^{m-n}.
4
Apply the square root to the simplified fraction, set it equal to 6464, and express both sides as powers of 2.
215x=215x2=64=26\sqrt{2^{15-x}} = 2^{\frac{15-x}{2}} = 64 = 2^6
The square root of a term is equivalent to raising that term to the power of 12\frac{1}{2}.
5
Equate the exponents and solve for xx.
15x2=6    15x=12    x=3\frac{15-x}{2} = 6 \implies 15-x = 12 \implies x = 3
If two exponential expressions with the same positive base are equal, their exponents must be equal.

Anahtar Kavram

Simplifying expressions using the laws of exponents and properties of roots
Soru 450Soru

A rectangular field has a length of 1.5×1041.5 \times 10^4 meters and a width of 4.0×1034.0 \times 10^3 meters. If a tractor can mow 6.0×1056.0 \times 10^5 square meters per hour, how many hours will it take the tractor to mow the entire field?

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Cevap: 100

Cevap

It will take 100 hours for the tractor to mow the entire field.
The total area of the field is the product of its length and width: (1.5×104)×(4.0×103)=6.0×107(1.5 \times 10^4) \times (4.0 \times 10^3) = 6.0 \times 10^7 square meters. Dividing this area by the tractor's mowing rate of 6.0×1056.0 \times 10^5 square meters per hour gives 6.0×1076.0×105=1.0×102=100\frac{6.0 \times 10^7}{6.0 \times 10^5} = 1.0 \times 10^2 = 100 hours.

Adım Adım Çözüm

1
Calculate the area of the rectangular field.
Area = 6.0×1076.0 \times 10^7 square meters
The area of a rectangle is found by multiplying its length by its width: (1.5×104 m)×(4.0×103 m)=6.0×107 m2(1.5 \times 10^4 \text{ m}) \times (4.0 \times 10^3 \text{ m}) = 6.0 \times 10^7 \text{ m}^2.
2
Calculate the time required to mow the field.
Time = 100100 hours
Divide the total area by the tractor's mowing rate: 6.0×107 m26.0×105 m2/hour=1.0×102=100\frac{6.0 \times 10^7 \text{ m}^2}{6.0 \times 10^5 \text{ m}^2/\text{hour}} = 1.0 \times 10^2 = 100 hours.

Anahtar Kavram

Multiplication and division of numbers in scientific notation
Soru 451Soru

A smart thermostat is programmed to reduce a home's heating energy usage. In January, the heating energy usage is reduced by 15%15\% compared to the baseline usage. In February, the usage is reduced by an additional 18\frac{1}{8} of January's usage level. In March, the usage is reduced by 20%20\% of February's usage level. If the baseline heating energy usage was 400400 kilowatt-hours (kWh), what is the total reduction in energy usage from the baseline to the end of March, in kWh?

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Cevap: 162

Cevap

The total reduction in energy usage from the baseline to the end of March is 162 kWh.
The baseline usage of 400400 kWh is reduced by 15%15\% in January, leaving 340340 kWh. In February, a reduction of 18\frac{1}{8} of 340340 kWh reduces usage by 42.542.5 kWh, leaving 297.5297.5 kWh. In March, a reduction of 20%20\% of 297.5297.5 kWh reduces usage by 59.559.5 kWh, leaving a final usage of 238238 kWh. The total reduction is the difference between the baseline and final usage levels, which is 162162 kWh.

Adım Adım Çözüm

1
Calculate January's energy reduction and January's usage level.
January reduction is 6060 kWh; January usage level is 340340 kWh.
January's reduction is 15%15\% of the 400400 kWh baseline (0.15×400=600.15 \times 400 = 60 kWh). Subtracting this reduction from the baseline gives January's usage level (40060=340400 - 60 = 340 kWh).
2
Calculate February's energy reduction and February's usage level.
February reduction is 42.542.5 kWh; February usage level is 297.5297.5 kWh.
February's reduction is 18\frac{1}{8} of January's usage level (340340 kWh), which is 0.125×340=42.50.125 \times 340 = 42.5 kWh. Subtracting this from January's level gives February's usage level (34042.5=297.5340 - 42.5 = 297.5 kWh).
3
Calculate March's energy reduction and March's usage level.
March reduction is 59.559.5 kWh; March usage level is 238238 kWh.
March's reduction is 20%20\% of February's usage level (297.5297.5 kWh), which is 0.20×297.5=59.50.20 \times 297.5 = 59.5 kWh. Subtracting this from February's level gives March's usage level (297.559.5=238297.5 - 59.5 = 238 kWh).
4
Calculate the total energy reduction from the baseline to the end of March.
The total reduction is 162162 kWh.
The total reduction can be found by adding the reductions from each of the three months (60+42.5+59.5=16260 + 42.5 + 59.5 = 162 kWh) or by subtracting the final usage level from the baseline (400238=162400 - 238 = 162 kWh).

Anahtar Kavram

Applying sequential percentage and fraction reductions to changing base values.
Soru 452Soru

On a vertical number line representing elevation in meters, a research drone is at coordinate dd and a submarine is at coordinate ss. The coordinate of the drone is a positive integer, and the coordinate of the submarine is a negative integer. The distance between the drone and the submarine is 150150 meters. If the absolute value of the submarine's coordinate is 44 times the coordinate of the drone, what is the coordinate of the drone?

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Cevap: 30

Cevap

The coordinate of the drone is 30.
The correct coordinate of 30 is found by defining the distance between the positive drone coordinate dd and negative submarine coordinate ss as ds=150d - s = 150. Since s<0s < 0, its absolute value s|s| is equal to s-s. Substituting s=4d-s = 4d gives d(4d)=150d - (-4d) = 150, which simplifies to 5d=1505d = 150 and yields d=30d = 30.

Adım Adım Çözüm

1
Set up equations based on the problem description.
d>0d > 0, s<0s < 0, distance ds=150|d - s| = 150, and s=4d|s| = 4d.
To translate the verbal description of elevations and distances into mathematical expressions.
2
Simplify the distance and absolute value expressions using the signs of the coordinates.
Since dd is positive and ss is negative, ds=150d - s = 150. Since ss is negative, s=s|s| = -s, so s=4d-s = 4d or s=4ds = -4d.
To eliminate absolute values based on the known signs of the variables.
3
Substitute the expression for ss into the distance equation and solve for dd.
d(4d)=1505d=150d=30d - (-4d) = 150 \Rightarrow 5d = 150 \Rightarrow d = 30.
To solve the system of linear equations to find the drone's coordinate.

Anahtar Kavram

Using absolute value to represent distance on a number line and solving equations involving sign constraints.
Soru 453Soru

The diameter of a human red blood cell is approximately 7.0×1067.0 \times 10^{-6} meters, and the diameter of a typical influenza virus is approximately 1.4×1071.4 \times 10^{-7} meters. How many times larger is the diameter of the red blood cell than the diameter of the influenza virus?

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Cevap: 50

Cevap

50
To determine how many times larger the diameter of the red blood cell is than the diameter of the influenza virus, divide the larger diameter by the smaller diameter: 7.0×1061.4×107\frac{7.0 \times 10^{-6}}{1.4 \times 10^{-7}}. Dividing the coefficients gives 7.01.4=5\frac{7.0}{1.4} = 5. Dividing the powers of ten using the quotient rule for exponents gives 106107=106(7)=101\frac{10^{-6}}{10^{-7}} = 10^{-6 - (-7)} = 10^1. Multiplying these results gives 5×101=505 \times 10^1 = 50.

Adım Adım Çözüm

1
Set up the ratio of the larger diameter to the smaller diameter
7.0×1061.4×107\frac{7.0 \times 10^{-6}}{1.4 \times 10^{-7}}
To find how many times larger one quantity is than another, divide the larger quantity by the smaller quantity.
2
Divide the decimal coefficients
5.05.0
Dividing 7.07.0 by 1.41.4 simplifies the numerical coefficient.
3
Divide the exponential terms using exponent properties
10110^1
Using the quotient rule for exponents, 10a10b=10ab\frac{10^a}{10^b} = 10^{a - b}, so 106(7)=10110^{-6 - (-7)} = 10^1.
4
Combine and simplify the final value
5050
Multiplying the coefficient by the simplified power of ten yields 5.0×10=505.0 \times 10 = 50.

Anahtar Kavram

Division of numbers in scientific notation using properties of exponents
Tahmini Süre:1m 30s
Soru 454Soru

A scientist monitors the temperature of two research chambers. Chamber A is kept at 5C-5^\circ\text{C} and Chamber B is kept at 7C7^\circ\text{C}. The scientist sets a third chamber, Chamber C, to a temperature of TCT^\circ\text{C} such that the distance between TT and the temperature of Chamber A on the Celsius scale is exactly 33 times the distance between TT and the temperature of Chamber B. If the temperature of Chamber C is warmer than Chamber A but colder than Chamber B, what is the value of TT?

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Cevap: 4

Cevap

The correct temperature value of Chamber C is 4.
The temperature of Chamber C must be 4C4^\circ\text{C} because the distance from 44 to 5-5 is 4(5)=9|4 - (-5)| = 9 and the distance from 44 to 77 is 47=3|4 - 7| = 3. The distance of 99 is exactly 33 times the distance of 33. Furthermore, 44 lies between 5-5 and 77, satisfying the condition that Chamber C is warmer than Chamber A but colder than Chamber B.

Adım Adım Çözüm

1
Represent the distances on the number line using absolute value expressions.
The distance to Chamber A is T(5)=T+5|T - (-5)| = |T + 5| and the distance to Chamber B is T7|T - 7|.
Distance between two points xx and yy on a number line is represented by xy|x - y|.
2
Set up the algebraic equation reflecting the relationship between the distances.
T+5=3T7|T + 5| = 3|T - 7|
The problem states the distance to Chamber A is exactly 3 times the distance to Chamber B.
3
Solve the absolute value equation by considering both positive and negative cases.
Case 1: T+5=3(T7)T=13T + 5 = 3(T - 7) \Rightarrow T = 13. Case 2: T+5=3(T7)T=4T + 5 = -3(T - 7) \Rightarrow T = 4.
The equation x=y|x| = |y| implies x=yx = y or x=yx = -y.
4
Verify which solution satisfies the temperature boundary condition.
Since Chamber C must be warmer than 5C-5^\circ\text{C} but colder than 7C7^\circ\text{C}, the only valid value is T=4T = 4.
The value T=13T = 13 is warmer than both chambers and does not lie between them.

Anahtar Kavram

Using absolute value to represent distance on a number line and solving absolute value equations with boundary conditions.

Alternatif Yöntem

Alternatively, visualize this on a number line. The total distance between Chamber A (5-5) and Chamber B (77) is 1212 units. Since Chamber C lies between them and the distance from C to A is 33 times the distance from C to B, we can divide the 1212-unit interval into 3+1=43 + 1 = 4 equal parts. Each part is 12÷4=312 \div 4 = 3 units. Chamber C is located 11 part away from Chamber B (towards Chamber A), which places it at 73=47 - 3 = 4.
Tahmini Süre:1m 30s
Soru 455Soru

A circle in the standard (x,y)(x, y) coordinate plane is defined by the equation x2+y28x6y=0x^2 + y^2 - 8x - 6y = 0. A point on this circle has an xx-coordinate of 11 and a positive yy-coordinate. What is the yy-coordinate of this point?

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Cevap: 7

Cevap

The positive y-coordinate of the point on the circle is 7.
Substituting x=1x = 1 into the circle's equation x2+y28x6y=0x^2 + y^2 - 8x - 6y = 0 gives 1+y286y=01 + y^2 - 8 - 6y = 0, which simplifies to the quadratic equation y26y7=0y^2 - 6y - 7 = 0. Factoring this equation yields (y7)(y+1)=0(y - 7)(y + 1) = 0. The solutions are y=7y = 7 and y=1y = -1. Because the y-coordinate must be positive, the correct value is 7.

Adım Adım Çözüm

1
Substitute the x-coordinate x=1x = 1 into the circle's equation.
12+y28(1)6y=01^2 + y^2 - 8(1) - 6y = 0
This sets up the equation to solve for the corresponding y-coordinates.
2
Simplify the equation.
y26y7=0y^2 - 6y - 7 = 0
Combining the constant terms creates a standard quadratic equation in terms of y.
3
Solve the quadratic equation by factoring.
(y7)(y+1)=0(y - 7)(y + 1) = 0, yielding y=7y = 7 or y=1y = -1
Factoring allows us to find the two possible y-values that satisfy the equation.
4
Choose the positive y-coordinate.
y=7y = 7
The question specifies that the y-coordinate must be positive.

Anahtar Kavram

Substituting a known coordinate into a circle's equation to find the other coordinate using quadratic equations.
Soru 456Soru

A highway is constructed up a mountain pass at a constant incline. At a distance of 3.63.6 miles from the starting toll gate, the elevation of the highway is 1,4201,420 feet above sea level. At a distance of 8.48.4 miles from the toll gate, the elevation is 3,1003,100 feet above sea level. What is the slope of the highway's elevation profile, in feet per mile?

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Cevap: 350

Cevap

The slope of the highway's elevation profile is 350 feet per mile.
The slope represents the constant rate of change of elevation per mile. By defining the coordinates as (x1,y1)=(3.6,1420)(x_1, y_1) = (3.6, 1420) and (x2,y2)=(8.4,3100)(x_2, y_2) = (8.4, 3100), we calculate the slope using the slope formula. Substituting the values gives m=310014208.43.6=16804.8=350m = \frac{3100 - 1420}{8.4 - 3.6} = \frac{1680}{4.8} = 350.

Adım Adım Çözüm

1
Represent the given data as coordinate points where the x-coordinate represents the distance in miles and the y-coordinate represents the elevation in feet.
The coordinate points are (3.6,1420)(3.6, 1420) and (8.4,3100)(8.4, 3100).
Slope is the change in the dependent variable (elevation) per unit change in the independent variable (distance).
2
Apply the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1} to find the rate of change.
m=310014208.43.6=16804.8m = \frac{3100 - 1420}{8.4 - 3.6} = \frac{1680}{4.8}
The slope represents the constant rate of change of elevation relative to the distance traveled.
3
Perform the division to find the numerical slope.
350
Dividing 1680 by 4.8 yields the exact value of 350.

Anahtar Kavram

The slope of a line represents a constant rate of change between two variables, calculated as the change in the vertical coordinate divided by the change in the horizontal coordinate.
Soru 457Soru

A manufacturer of solar panels inspects their production in three stages. In the first stage, 112\frac{1}{12} of the total panels produced are identified as defective and recycled. In the second stage, 20%20\% of the remaining non-defective panels are found to have cosmetic flaws and are sold at a discount. In the third stage, 15\frac{1}{5} of the panels that passed the first two stages without any defects or flaws are selected for quality testing. If 1,7601,760 panels passed the first two stages but were NOT selected for quality testing, what was the total number of panels originally produced?

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Cevap: 3000

Cevap

The total number of panels originally produced was 3,0003,000.
The correct answer of 3,0003,000 is determined by calculating the cumulative fraction of panels that are not defective, do not have cosmetic flaws, and are not selected for testing. In Stage 1, subtracting 112\frac{1}{12} leaves 1112\frac{11}{12} of the original panels. In Stage 2, subtracting 20%20\% of those leaves 80%80\% (or 45\frac{4}{5}), resulting in 1115\frac{11}{15} of the original panels. In Stage 3, subtracting 15\frac{1}{5} of these leaves 45\frac{4}{5}, resulting in 4475\frac{44}{75} of the original panels. Setting 4475\frac{44}{75} of the original panels equal to 1,7601,760 and solving for the total original panels yields 3,0003,000.

Adım Adım Çözüm

1
Calculate the fraction of panels remaining after the first stage of inspection.
1112\frac{11}{12} of the original panels
Since 112\frac{1}{12} of the total panels are defective and recycled, the remaining portion is 1112=11121 - \frac{1}{12} = \frac{11}{12}.
2
Calculate the fraction of the original panels that pass the second stage without defects or flaws.
1115\frac{11}{15} of the original panels
Of the remaining 1112\frac{11}{12} of the panels, 20%20\% have cosmetic flaws, meaning 100%20%=80%100\% - 20\% = 80\% (or 45\frac{4}{5}) do not have flaws. Multiplying these fractions gives 45×1112=1115\frac{4}{5} \times \frac{11}{12} = \frac{11}{15}.
3
Calculate the fraction of original panels that pass the first two stages but are not selected for testing.
4475\frac{44}{75} of the original panels
Since 15\frac{1}{5} of the flawless panels are selected for testing, the remaining 115=451 - \frac{1}{5} = \frac{4}{5} are not selected. Multiplying this by the fraction from Step 2 gives 45×1115=4475\frac{4}{5} \times \frac{11}{15} = \frac{44}{75}.
4
Solve for the original number of panels, PP, using the given value of 1,7601,760.
P=3,000P = 3,000 panels
Set up the equation 4475P=1,760\frac{44}{75} P = 1,760. Solving for PP gives P=1,760×7544=40×75=3,000P = 1,760 \times \frac{75}{44} = 40 \times 75 = 3,000.

Anahtar Kavram

Applying successive fractions and percentages to resolve a multi-stage word problem by working backward to the initial value.
Soru 458Soru

What is the exact value of cos(arcsin(35))\cos\left(\arcsin\left(\frac{3}{5}\right)\right) expressed as a decimal?

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Cevap: 0.8

Cevap

The exact value of the expression is 0.8.
Evaluating cos(arcsin(35))\cos\left(\arcsin\left(\frac{3}{5}\right)\right) requires finding the cosine of an angle θ\theta whose sine is 35\frac{3}{5}. In a right triangle with an opposite side of 3 and a hypotenuse of 5, the adjacent side is 5232=4\sqrt{5^2 - 3^2} = 4. The cosine of θ\theta is the ratio of the adjacent side to the hypotenuse, which gives 45=0.8\frac{4}{5} = 0.8.

Adım Adım Çözüm

1
Interpret the inverse sine function as an angle.
Let θ=arcsin(35)\theta = \arcsin\left(\frac{3}{5}\right), meaning sin(θ)=35\sin(\theta) = \frac{3}{5} for 0<θ<π20 < \theta < \frac{\pi}{2}.
The inverse sine function returns an angle whose sine is the given value within the principal interval [π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].
2
Find the adjacent side of the right triangle associated with angle θ\theta.
adjacent=5232=16=4\text{adjacent} = \sqrt{5^2 - 3^2} = \sqrt{16} = 4.
By the Pythagorean theorem, a2+b2=c2a^2 + b^2 = c^2, so the adjacent side length is c2b2\sqrt{c^2 - b^2}.
3
Calculate the cosine of angle θ\theta.
cos(θ)=adjacenthypotenuse=45=0.8\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5} = 0.8.
Cosine is defined as the ratio of the adjacent side to the hypotenuse in a right triangle.

Anahtar Kavram

Composition of Trigonometric and Inverse Trigonometric Functions
Tahmini Süre:45s
Soru 459Soru

In rectangle ABCDABCD, the length of side ADAD is 1212 units and the length of side CDCD is 1717 units. Point EE lies on side CDCD such that ADE\triangle ADE is an isosceles right triangle with the right angle at vertex DD. What is the length, in units, of segment BEBE?

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Cevap: 13

Cevap

The length of segment BEBE is 1313 units.
Because ADE\triangle ADE is an isosceles right triangle with the right angle at vertex DD, leg DEDE equals leg AD=12AD = 12. Subtracting DEDE from total side length CD=17CD = 17 gives segment EC=5EC = 5. Since ABCDABCD is a rectangle, angle CC is a right angle (9090^\circ) and BC=AD=12BC = AD = 12. Applying the Pythagorean Theorem to right triangle BCE\triangle BCE gives hypotenuse BE=122+52=144+25=169=13BE = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13.

Adım Adım Çözüm

1
Find the length of segment DEDE using the properties of an isosceles right triangle.
DE=12DE = 12 units
In isosceles right triangle ADE\triangle ADE with right angle at DD, legs ADAD and DEDE are equal in length. Given AD=12AD = 12, DEDE must also be 1212.
2
Determine the length of segment ECEC.
EC=5EC = 5 units
Since point EE lies on side CDCD, EC=CDDE=1712=5EC = CD - DE = 17 - 12 = 5.
3
Use the Pythagorean Theorem in right triangle BCE\triangle BCE to find BEBE.
BE=13BE = 13 units
Because ABCDABCD is a rectangle, angle CC is 9090^\circ and BC=AD=12BC = AD = 12. Applying the Pythagorean Theorem with legs BC=12BC = 12 and EC=5EC = 5 gives BE=122+52=169=13BE = \sqrt{12^2 + 5^2} = \sqrt{169} = 13.

Anahtar Kavram

Applying properties of isosceles right triangles (45459045^\circ-45^\circ-90^\circ) and the Pythagorean Theorem in composite figures.
Soru 460Soru

If sinθcosθ=0.6\sin \theta - \cos \theta = 0.6, what is the value of sinθcosθ\sin \theta \cos \theta?

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Cevap: 0.32

Cevap

The value of sinθcosθ\sin \theta \cos \theta is 0.32.
Squaring both sides of sinθcoscosθ=0.6\sin \theta - \cos \cos \theta = 0.6 yields sin2θ2sinθcosθ+cos2θ=0.36\sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta = 0.36. Substituting the Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 gives 12sinθcosθ=0.361 - 2\sin \theta \cos \theta = 0.36, which rearranges to 2sinθcosθ=0.642\sin \theta \cos \theta = 0.64. Dividing by 2 yields sinθcosθ=0.32\sin \theta \cos \theta = 0.32.

Adım Adım Çözüm

1
Square both sides of the given equation
(sinθcosθ)2=0.36(\sin \theta - \cos \theta)^2 = 0.36
Squaring allows us to introduce the product sinθcosθ\sin \theta \cos \theta alongside sin2θ\sin^2 \theta and cos2θ\cos^2 \theta.
2
Expand the binomial on the left side
sin2θ2sinθcosθ+cos2θ=0.36\sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta = 0.36
Use the algebraic expansion identity (ab)2=a22ab+b2(a - b)^2 = a^2 - 2ab + b^2.
3
Substitute the fundamental Pythagorean trigonometric identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1
12sinθcosθ=0.361 - 2\sin \theta \cos \theta = 0.36
sin2θ+cos2θ\sin^2 \theta + \cos^2 \theta always equals 1 for any angle θ\theta.
4
Isolate the term containing sinθcosθ\sin \theta \cos \theta
2sinθcosθ=0.642\sin \theta \cos \theta = 0.64
Subtract 0.36 from 1 to find the value of 2sinθcosθ2\sin \theta \cos \theta.
5
Divide by 2 to solve for sinθcosθ\sin \theta \cos \theta
sinθcosθ=0.32\sin \theta \cos \theta = 0.32
Simplifies 0.64/20.64 / 2 to obtain the final required numerical value.

Anahtar Kavram

Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1
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