3(x-6)<(x-1)
3x - 18 < x - 1 (distribute the 3)
3x - x < 18 - 1 (move x to the left and 1 to the right)
2x < 17
x < 17/2
The result of the inequality is x is less than 17/2.
The result from the solving of this inequality 3(x-6)<(x-1) is
39 answers
A ship sails from p to Q on a bearing of 60°,and then from Q to R on a bearing of 300°. PQ=8km and QR =5km.find
a: How far is the ship now from its starting point?
b:The appropriate bearing of R from 1 is
a: How far is the ship now from its starting point?
b:The appropriate bearing of R from 1 is
To solve this problem, we can use the Law of Cosines and the Law of Sines.
a) To find how far the ship is now from its starting point, we need to find the length of PR. Using the Law of Cosines:
PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)cos(Q)
PR^2 = 8^2 + 5^2 - 2(8)(5)cos(180° - 60°)
PR^2 = 64 + 25 - 80cos(120°)
PR^2 = 89 - 80(-0.5)
PR^2 = 89 + 40
PR^2 = 129
PR ≈ √129
Therefore, the ship is now approximately √129 km from its starting point.
b) To find the appropriate bearing of R from 1, we can use the Law of Sines. Let angle P be θ.
sin(θ)/PR = sin(60°)/QR
sin(θ)/(√129) = sin(60°)/5
Cross multiply:
5sin(θ) = (√129)sin(60°)
Divide by 5:
sin(θ) = (√129/5)sin(60°)
Take the arcsin (inverse sine) of both sides:
θ ≈ arcsin(√129/5 * sin(60°))
Use a calculator to find:
θ ≈ arcsin(√129/5 * 0.866)
θ ≈ arcsin(0.866 * 7.16)
θ ≈ arcsin(6.20)
Therefore, the appropriate bearing of R from 1 is approximately arcsin(6.20).
a) To find how far the ship is now from its starting point, we need to find the length of PR. Using the Law of Cosines:
PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)cos(Q)
PR^2 = 8^2 + 5^2 - 2(8)(5)cos(180° - 60°)
PR^2 = 64 + 25 - 80cos(120°)
PR^2 = 89 - 80(-0.5)
PR^2 = 89 + 40
PR^2 = 129
PR ≈ √129
Therefore, the ship is now approximately √129 km from its starting point.
b) To find the appropriate bearing of R from 1, we can use the Law of Sines. Let angle P be θ.
sin(θ)/PR = sin(60°)/QR
sin(θ)/(√129) = sin(60°)/5
Cross multiply:
5sin(θ) = (√129)sin(60°)
Divide by 5:
sin(θ) = (√129/5)sin(60°)
Take the arcsin (inverse sine) of both sides:
θ ≈ arcsin(√129/5 * sin(60°))
Use a calculator to find:
θ ≈ arcsin(√129/5 * 0.866)
θ ≈ arcsin(0.866 * 7.16)
θ ≈ arcsin(6.20)
Therefore, the appropriate bearing of R from 1 is approximately arcsin(6.20).
2+2
2 + 2 = 4
The mean of a set of number is 68,65, x, 69, 77,48,64 and 67. find x
To find the value of x, we need to calculate the mean of the given set of numbers and set it equal to 68.
Mean = (sum of all numbers) / (number of numbers)
Therefore, we can set up the equation:
(68 + 65 + x + 69 + 77 + 48 + 64 + 67) / 8 = 68
Now, let's solve for x:
(351 + x) / 8 = 68
Multiply both sides by 8:
351 + x = 544
Subtract 351 from both sides:
x = 544 - 351
x = 193
Therefore, the value of x is 193.
Mean = (sum of all numbers) / (number of numbers)
Therefore, we can set up the equation:
(68 + 65 + x + 69 + 77 + 48 + 64 + 67) / 8 = 68
Now, let's solve for x:
(351 + x) / 8 = 68
Multiply both sides by 8:
351 + x = 544
Subtract 351 from both sides:
x = 544 - 351
x = 193
Therefore, the value of x is 193.
The average of 10 boys was 12 years.A boy of 15 years was replaced with that of 5 years.Find the new average age of the boys
The total age of the 10 boys before the replacement is 10 * 12 = 120 years.
To find the new average age, we need to consider the sum of their ages after the replacement.
Since a boy of 15 years was replaced with a boy of 5 years, the total age difference is 15 - 5 = 10 years.
The new total age of the boys is 120 - 10 = 110 years.
To find the new average age, we divide the new total age by the number of boys (which is still 10):
New average age = 110 / 10 = 11 years.
Therefore, the new average age of the boys is 11 years after the replacement.
To find the new average age, we need to consider the sum of their ages after the replacement.
Since a boy of 15 years was replaced with a boy of 5 years, the total age difference is 15 - 5 = 10 years.
The new total age of the boys is 120 - 10 = 110 years.
To find the new average age, we divide the new total age by the number of boys (which is still 10):
New average age = 110 / 10 = 11 years.
Therefore, the new average age of the boys is 11 years after the replacement.
Twenty girls and boys sat for an examination.The mean marks obtained by the girls and boys were 62 and 57 respectively.if the total scores for both girls and boys were 2950, find Y
Let y represent the number of boys in the group.
Since there are 20 students total, the number of girls would be 20 - y.
The total marks obtained by the girls would be the mean marks (62) multiplied by the number of girls.
Similarly, the total marks obtained by the boys would be the mean marks (57) multiplied by the number of boys.
So, we can form the following equations:
62(20 - y) + 57y = 2950
Expanding the equation:
1240 - 62y + 57y = 2950
Combining like terms:
-5y = 170
Dividing both sides by -5:
y = -34
Since we cannot have a negative number of boys, this solution does not make sense in the context of the problem.
Therefore, there may have been an error in the given information or calculations. Please double-check and provide the correct information for further assistance.
Since there are 20 students total, the number of girls would be 20 - y.
The total marks obtained by the girls would be the mean marks (62) multiplied by the number of girls.
Similarly, the total marks obtained by the boys would be the mean marks (57) multiplied by the number of boys.
So, we can form the following equations:
62(20 - y) + 57y = 2950
Expanding the equation:
1240 - 62y + 57y = 2950
Combining like terms:
-5y = 170
Dividing both sides by -5:
y = -34
Since we cannot have a negative number of boys, this solution does not make sense in the context of the problem.
Therefore, there may have been an error in the given information or calculations. Please double-check and provide the correct information for further assistance.
The age of abu, segun, kofi, and fommi are 17 years (2x-13) years, 14 years and 16 years respectively. what is the value of x if their mean age is 17.5 years
To find the value of x, we need to calculate the mean age of Abu, Segun, Kofi, and Fommi and set it equal to 17.5 years.
Mean Age = (Sum of all ages) / (Number of individuals)
Using the given information, the sum of their ages can be expressed as:
17 + (2x-13) + 14 + 16
The sum of their ages is then:
47 + 2x - 13
To find the mean age, we divide this sum by the number of individuals (4):
[47 + 2x - 13] / 4 = 17.5
Now, let's solve for x:
47 + 2x - 13 = 17.5 * 4
34 + 2x = 70
Subtract 34 from both sides:
2x = 36
Divide both sides by 2:
x = 18
Therefore, the value of x is 18.
Mean Age = (Sum of all ages) / (Number of individuals)
Using the given information, the sum of their ages can be expressed as:
17 + (2x-13) + 14 + 16
The sum of their ages is then:
47 + 2x - 13
To find the mean age, we divide this sum by the number of individuals (4):
[47 + 2x - 13] / 4 = 17.5
Now, let's solve for x:
47 + 2x - 13 = 17.5 * 4
34 + 2x = 70
Subtract 34 from both sides:
2x = 36
Divide both sides by 2:
x = 18
Therefore, the value of x is 18.
The weight of 5 girls in kg are 48,x,52,50 and (2x-5).if their average weight is 47kg.find the weight of the girl that has the heaviest weight
To find the weight of the girl with the heaviest weight, we need to determine the value of x.
To do this, we can use the average weight formula:
Average = Sum of all weights / Number of girls
Given that the average weight is 47 kg, we can set up the equation:
47 = (48 + x + 52 + 50 + (2x-5)) / 5
Multiplying both sides by 5:
235 = 48 + x + 52 + 50 + (2x-5)
Combining like terms:
235 = 155 + 3x
Subtracting 155 from both sides:
80 = 3x
Dividing both sides by 3:
x = 80/3
x ≈ 26.67
Now that we have the value of x, we can substitute it back into the equation to find the weight of the girl with the heaviest weight:
Weight of girl = 2x - 5
Weight of girl = 2(26.67) - 5
Weight of girl ≈ 53.34 - 5
Weight of girl ≈ 48.34 kg
Therefore, the girl with the heaviest weight weighs approximately 48.34 kg.
To do this, we can use the average weight formula:
Average = Sum of all weights / Number of girls
Given that the average weight is 47 kg, we can set up the equation:
47 = (48 + x + 52 + 50 + (2x-5)) / 5
Multiplying both sides by 5:
235 = 48 + x + 52 + 50 + (2x-5)
Combining like terms:
235 = 155 + 3x
Subtracting 155 from both sides:
80 = 3x
Dividing both sides by 3:
x = 80/3
x ≈ 26.67
Now that we have the value of x, we can substitute it back into the equation to find the weight of the girl with the heaviest weight:
Weight of girl = 2x - 5
Weight of girl = 2(26.67) - 5
Weight of girl ≈ 53.34 - 5
Weight of girl ≈ 48.34 kg
Therefore, the girl with the heaviest weight weighs approximately 48.34 kg.
in Angle pqr. P=150 °, p= 12cm, q =8cm. sin Q is
To find sin Q, we can use the law of sines, which states that in a triangle, the ratio of the length of a side to the sine of its opposite angle is constant.
The law of sines can be written as:
sin P / p = sin Q / q
Given that P = 150°, p = 12 cm, and q = 8 cm, we can substitute these values into the law of sines:
sin 150° / 12 = sin Q / 8
We know that sin 150° is equal to sin (180° - 150°) since the sine function is periodic:
sin 150° = sin 30°
Therefore, we can rewrite the equation as:
sin 30° / 12 = sin Q / 8
Simplifying further:
1/2 / 12 = sin Q / 8
1/24 = sin Q / 8
To isolate sin Q, we can cross-multiply:
sin Q = (1/24) * 8
sin Q = 1/3
Therefore, sin Q is equal to 1/3.
The law of sines can be written as:
sin P / p = sin Q / q
Given that P = 150°, p = 12 cm, and q = 8 cm, we can substitute these values into the law of sines:
sin 150° / 12 = sin Q / 8
We know that sin 150° is equal to sin (180° - 150°) since the sine function is periodic:
sin 150° = sin 30°
Therefore, we can rewrite the equation as:
sin 30° / 12 = sin Q / 8
Simplifying further:
1/2 / 12 = sin Q / 8
1/24 = sin Q / 8
To isolate sin Q, we can cross-multiply:
sin Q = (1/24) * 8
sin Q = 1/3
Therefore, sin Q is equal to 1/3.
Given that the bearing of P from Q is 115°, the bearing of Q from P is
To find the bearing of Q from P, we need to determine the angle formed between the line joining P and Q and the north direction.
The bearing of Q from P is the angle measured in a clockwise direction from the north line to the line joining P and Q.
Since the bearing of P from Q is given as 115°, the bearing of Q from P will be the supplementary angle to 115°.
Supplementary angles add up to 180°.
Therefore, the bearing of Q from P is:
180° - 115° = 65°
The bearing of Q from P is 65°.
The bearing of Q from P is the angle measured in a clockwise direction from the north line to the line joining P and Q.
Since the bearing of P from Q is given as 115°, the bearing of Q from P will be the supplementary angle to 115°.
Supplementary angles add up to 180°.
Therefore, the bearing of Q from P is:
180° - 115° = 65°
The bearing of Q from P is 65°.
in the diagram below, AI=4cm,IB=10cm and ID=8cm.find the length of IC
Since AI and IB form a right angle at point I, we can use the Pythagorean theorem to find the length of IC.
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, IC is the hypotenuse, and AI and AB are the other two sides.
So, applying the Pythagorean theorem:
IC^2 = AI^2 + AB^2
IC^2 = 4^2 + 10^2
IC^2 = 16 + 100
IC^2 = 116
Taking the square root of both sides:
IC = √116
IC ≈ 10.77 cm
Therefore, the length of IC is approximately 10.77 cm.
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, IC is the hypotenuse, and AI and AB are the other two sides.
So, applying the Pythagorean theorem:
IC^2 = AI^2 + AB^2
IC^2 = 4^2 + 10^2
IC^2 = 16 + 100
IC^2 = 116
Taking the square root of both sides:
IC = √116
IC ≈ 10.77 cm
Therefore, the length of IC is approximately 10.77 cm.
The smallest integer that can satisfy the inequality 30-5x <(x-1) is
To solve the inequality 30 - 5x < (x - 1), we will follow these steps:
1. Simplify both sides of the inequality:
30 - 5x < x - 1
2. Move all x terms to one side and constant terms to the other:
-5x - x < -1 - 30
-6x < -31
3. Divide both sides by -6. Remember to reverse the inequality sign when dividing by a negative number:
x > (-31)/-6
Simplifying further:
x > 31/6
The smallest integer that satisfies the inequality x > 31/6 is 6 (since 31/6 ≈ 5.16). However, the inequality is strict (>) and not inclusive of x = 6, so the smallest integer that satisfies the inequality is the next integer greater than 6, which is 7.
Therefore, the smallest integer that can satisfy the inequality 30 - 5x < (x - 1) is 7.
1. Simplify both sides of the inequality:
30 - 5x < x - 1
2. Move all x terms to one side and constant terms to the other:
-5x - x < -1 - 30
-6x < -31
3. Divide both sides by -6. Remember to reverse the inequality sign when dividing by a negative number:
x > (-31)/-6
Simplifying further:
x > 31/6
The smallest integer that satisfies the inequality x > 31/6 is 6 (since 31/6 ≈ 5.16). However, the inequality is strict (>) and not inclusive of x = 6, so the smallest integer that satisfies the inequality is the next integer greater than 6, which is 7.
Therefore, the smallest integer that can satisfy the inequality 30 - 5x < (x - 1) is 7.
the smallest integer that can satisfy the inequality 30-5x < 2x ±-2 is
To solve the inequality 30-5x < 2x ± 2, we will follow these steps:
1. Simplify both sides of the inequality:
30 - 5x < 2x ± 2
2. We'll consider the two cases separately: one with the positive sign (2x + 2) and the other with the negative sign (2x - 2).
For 2x + 2:
30 - 5x < 2x + 2
3. Move all x terms to one side and constant terms to the other:
-5x - 2x < 2 - 30
-7x < -28
4. Divide both sides by -7. Remember to reverse the inequality sign when dividing by a negative number:
x > (-28)/-7
x > 4
For 2x - 2:
30 - 5x < 2x - 2
3. Move all x terms to one side and constant terms to the other:
-2 - 5x < 2x - 30
4. Combine like terms:
5x + 2x < 30 - 2
5. Simplify further:
7x < 28
6. Divide both sides by 7:
x < 28/7
x < 4
Therefore, x is less than 4.
To find the smallest integer that satisfies both conditions, x should be greater than 4 (from 2x + 2) and less than 4 (from 2x - 2). However, this is not possible since there are no integers that are both greater than 4 and less than 4.
Thus, there is no smallest integer that can satisfy the inequality 30 - 5x < 2x ± 2.
1. Simplify both sides of the inequality:
30 - 5x < 2x ± 2
2. We'll consider the two cases separately: one with the positive sign (2x + 2) and the other with the negative sign (2x - 2).
For 2x + 2:
30 - 5x < 2x + 2
3. Move all x terms to one side and constant terms to the other:
-5x - 2x < 2 - 30
-7x < -28
4. Divide both sides by -7. Remember to reverse the inequality sign when dividing by a negative number:
x > (-28)/-7
x > 4
For 2x - 2:
30 - 5x < 2x - 2
3. Move all x terms to one side and constant terms to the other:
-2 - 5x < 2x - 30
4. Combine like terms:
5x + 2x < 30 - 2
5. Simplify further:
7x < 28
6. Divide both sides by 7:
x < 28/7
x < 4
Therefore, x is less than 4.
To find the smallest integer that satisfies both conditions, x should be greater than 4 (from 2x + 2) and less than 4 (from 2x - 2). However, this is not possible since there are no integers that are both greater than 4 and less than 4.
Thus, there is no smallest integer that can satisfy the inequality 30 - 5x < 2x ± 2.
in the diagram below O is the centre of the circle, find the value of x
I'm sorry, but I cannot see the diagram you are referring to. Can you please provide more information or describe the diagram to me?
in the diagram below O is centre of the circle RT is a diameter <PQT=33° and <TO=76. calculate the value of angle PTR and find the size of angle PRS
Since RT is a diameter, angle PTR is a right angle (90°) because it is an angle formed by a line (RT) that passes through the center of the circle (O).
To find the size of angle PRS, we need to calculate the measure of angle PTS. Since angle PQT is given as 33°, we can use the fact that angles on a straight line add up to 180°.
Angle PTS = 180° - Angle PQT
Angle PTS = 180° - 33°
Angle PTS = 147°
Now, in triangle PTS, we can use the fact that the angles of a triangle add up to 180°:
Angle PRS + Angle PTS + Angle PTR = 180°
Substituting the known values:
Angle PRS + 147° + 90° = 180°
Simplifying the equation:
Angle PRS + 237° = 180°
Subtracting 237° from both sides:
Angle PRS = -57°
Since the result is negative, it means there must have been an error in the given information or calculations. Please double-check and provide the correct information for further assistance.
To find the size of angle PRS, we need to calculate the measure of angle PTS. Since angle PQT is given as 33°, we can use the fact that angles on a straight line add up to 180°.
Angle PTS = 180° - Angle PQT
Angle PTS = 180° - 33°
Angle PTS = 147°
Now, in triangle PTS, we can use the fact that the angles of a triangle add up to 180°:
Angle PRS + Angle PTS + Angle PTR = 180°
Substituting the known values:
Angle PRS + 147° + 90° = 180°
Simplifying the equation:
Angle PRS + 237° = 180°
Subtracting 237° from both sides:
Angle PRS = -57°
Since the result is negative, it means there must have been an error in the given information or calculations. Please double-check and provide the correct information for further assistance.
in the diagram below O is centre of the circle RT is a diameter <PQT =33° and <TO =76°. calculate the value of angle PTR
To find the value of angle PTR, we need to split it into two smaller angles: angle PTO and angle RTO.
Since angle PQT is given as 33°, and angle TO is given as 76°, we can determine angle PTO and angle RTO.
Angle PTO is formed by the radius OP and the tangent line PT, so it is 90 degrees.
To find angle RTO, we can use the fact that angles in a triangle add up to 180 degrees.
In triangle RTO, we have angle RTO + angle TOR + angle ORT = 180°.
Since angle TOR is 90° (angle PTO) and angle ORT is half of angle TO (76°/2 = 38°), we can substitute these values into the equation:
Angle RTO + 90° + 38° = 180°
Simplifying the equation:
Angle RTO + 128° = 180°
Subtracting 128° from both sides:
Angle RTO = 52°
Now, we can find angle PTR by adding angle PTO (90°) and angle RTO (52°):
Angle PTR = 90° + 52°
Angle PTR = 142°
Therefore, the value of angle PTR is 142°.
Since angle PQT is given as 33°, and angle TO is given as 76°, we can determine angle PTO and angle RTO.
Angle PTO is formed by the radius OP and the tangent line PT, so it is 90 degrees.
To find angle RTO, we can use the fact that angles in a triangle add up to 180 degrees.
In triangle RTO, we have angle RTO + angle TOR + angle ORT = 180°.
Since angle TOR is 90° (angle PTO) and angle ORT is half of angle TO (76°/2 = 38°), we can substitute these values into the equation:
Angle RTO + 90° + 38° = 180°
Simplifying the equation:
Angle RTO + 128° = 180°
Subtracting 128° from both sides:
Angle RTO = 52°
Now, we can find angle PTR by adding angle PTO (90°) and angle RTO (52°):
Angle PTR = 90° + 52°
Angle PTR = 142°
Therefore, the value of angle PTR is 142°.
The bearing S40°E is the same as
The bearing S40°E is the same as 180° - 40°, which simplifies to S140°E.
Therefore, the bearing S40°E is the same as S140°E.
Therefore, the bearing S40°E is the same as S140°E.
The bearing of S50°W is the same as
The bearing of S50°W is the same as 180° + 50°, which simplifies to S230°W.
Therefore, the bearing of S50°W is the same as S230°W.
Therefore, the bearing of S50°W is the same as S230°W.
The diagram above MP is a diameter of the circle MQ is a straight, <NMP=42°,<NQR=X and NQ is a tangent to the circle at N. calculate the value of x
To find the value of x, we need to use the properties of tangents and angles within circles.
Since MP is a diameter of the circle, angle NMP is a right angle (90°) because it is an angle formed by a line (MP) that passes through the center of the circle.
Since NQ is a tangent to the circle at N, the angle formed between the tangent and the radius at the point of tangency is always 90°.
Therefore, angle NQR (denoted as x) is 90°.
So, the value of x is 90°.
Since MP is a diameter of the circle, angle NMP is a right angle (90°) because it is an angle formed by a line (MP) that passes through the center of the circle.
Since NQ is a tangent to the circle at N, the angle formed between the tangent and the radius at the point of tangency is always 90°.
Therefore, angle NQR (denoted as x) is 90°.
So, the value of x is 90°.
O is the centre of the circle, WY and XZ are straight line, < WP2=102° <XWY=30° and <PYZ=M. find the value of m
To find the value of angle M, we need to use the properties of angles formed by intersecting lines and angles within circles.
First,
First,