Forum Replies Created
-
AuthorPosts
-
in reply to: Substitution #18964
The sum of a geometric sequence is: $$S=\frac{a(1-r^n)}{1-r}$$. Find a when r = 0.8, n = 5 and S = 10. $$10=\frac{a(1-0.8^5)}{1-0.8}$$ substitute all numbers into their corresponding letters $$10=\frac{a(1-0.8^5)}{0.2}$$ simplify where possible $$10\times0.2=\frac{a(1-0.8^5)}{\cancel{0.2}}^{\times\cancel{0.2}}$$ multiply both sides by 0.2 2 = a(1 – 0.85) divide both sides by (1 – 0.85) $$a=\frac{2}{1-0.8^5}$$ calculate a = 2.97 in reply to: Substitution #18963If $$I=\frac{E}{R+r}$$. find r if I = 12, E = 60 and R = 4. $$12=\frac{60}{4+r}$$ substitute all numbers into their corresponding letters $$12(4+r)=\frac{60}{\cancel{(4+r)}}^{\times\cancel{(4+r)}}$$ multiply both sides by 4 + r 48 + 12r = 60 expand 12r = 12 subtract 48 from both sides r = 1 divide both sides by 12 Simplify 2(x – 1) – 3(x – 3) we must be especially careful when expanding with a negative sign = 2x – 1 – 3x + 9 2(x – 1) = 2 × × + 2 × -1
= 2x – 2
-3(x – 3) = -3 × × + -3 × -3
= -3x + 9
= –x + 8Only add and subtract like terms, the x’s with the x’s and the numbers with the numbers, paying extra special attention to the negative signs
2x – 3x = –x
-1 + 9 = 8Simplify ##\frac{{6{x^{\frac{3}{2}}}{y^{\frac{1}{2}}} \times {x^{\frac{4}{5}y\frac{3}{5}}}}}{{2{{\left( {{x^{\frac{1}{2}}}y} \right)}^{\frac{1}{5}}} \times 3{x^{\frac{1}{2}}}{y^{\frac{1}{5}}}}}##
## = \frac{{6{x^{\frac{{23}}{{10}}}}{y^{\frac{{11}}{{10}}}}}}{{6{x^{\frac{1}{{10}}}}{y^{\frac{1}{5}}} \times {x^{\frac{1}{2}}}{y^{\frac{1}{5}}}}}##
## = \frac{{{x^{\frac{{23}}{{10}}}}{y^{\frac{{11}}{{10}}}}}}{{{x^{\frac{3}{5}}}{y^{\frac{2}{5}}}}}##
## = {x^{\frac{{17}}{{10}}}}{y^{\frac{7}{{10}}}}##
Evaluate 49½ + 50– 2-1,showing all steps of your working. $$49^{\frac12}=\sqrt{49}$$
= 7
49 to the power of a half means the square root of 49 which is 7 $$5^0=1$$
anything to the power of zero is 1, so 5 to the power of zero is 1 $$2^{-1}-\frac12$$
2 to the power of negative 1 means 1 over 2 to the power of 1, which is the same as 1 over 2 $$49^{\frac12}+5^0-2^{-1}-\frac12$$
$$=\sqrt{49}+1-\frac12$$
=$$7\frac12$$
in reply to: Changing The Subject #18956Make s the subject of v² = u² + 2as.
v² – u² = 2as
subtract u2 from both sides ##\frac{v^2-u^2}{2a}=\frac{\cancel{2a}s}{\cancel{2a}}## divide both sides by 2a ##s=\frac{v^2-u^2}{2a}## in reply to: Changing The Subject #18954The heat (Q) required to raise the temperature of a steel rod, of mass (m), from a temperature of A to a temperature of (B) is given by Q = 5m(B – A). Rearrange formula to make A the subject. Q = 5m(B – A) we are trying to get A by itself, so we must remove all the other letters and terms. Start by dividing both sides by 5m ##\frac{Q}{5m}=B-A## Add A to both sides so that we can get rid of the negative sign ##\frac{Q}{5m}+A=B## subtract $$\frac{Q}{5m}$$ from both sides to leave A by itself ##A= B-\frac{Q}{5m}## we could combine the fraction with the lowest common denominator ##A=\frac{5mB-Q}{5m}## in reply to: Changing The Subject #18953Change the subject of y = mx + b to x. y = mx + b we are trying to get x by itself, so we must remove all the other letters and terms. Start by subtracting b from both sides y – b = mx
Divide both sides by m ##\frac{y-b}{m}=x##
swap sides so the x is on the left ##x=\frac{y-b}{m}## Solve for x: ##\frac{{3x}}{4} – 5 = \frac{x}{5} + \frac{1}{2}##
##^{\times\cancel4\times5\times2}\frac{{3x}}{\cancel4} – 5^{\times4\times5\times2} = \frac{x}{\cancel5}^{\times4\times\cancel5\times2} + \frac{1}{\cancel2}^{\times4\times5\times\cancel2}## multiply everything by 4, 5 and 2 30x – 200 = 8x + 20
simplify 22x – 200 = 20
subtract 8x from both sides 22x = 220
add 200 to both sides x = 10 divide both sides by 22 If 1.05x = 2. Find the value of x. x = 2
1.052 = 1.1025Method 1: Guess and Check you start with any number and check, then modify the next guess until you get close to the answer 1.0510 = 1.62
1.0520 = 2.65
1.0515 = 2.07
1.0514 = 1.97
1.0514.5 = 2.02
1.0514.2 = 1.999
∴ x = 14.2Start with any number. We are starting with x = 2
Calculate and see how close you are to the answer you want
since it is too small, try a bigger number, say x = 10, which brings us closer to the answer, but we are still too small
x = 20 was too big, so we now know it is between 10 and 20.
Keep trying and refine your estimate until you get to the answer$$x=\frac{log\;2}{log\;1.05}$$
x = 14.2
Method 2: Using logs you don’t have to understand the log here, just follow the rule and memorise it. This is a more advanced method, but will always work for these types of equations if you can just remember which one is on top and which one is on the bottom
x= log the answer part (2) divided by log the question part (1.05)in reply to: Stopping Distance, Distance, Speed and Time #18946Maryam is driving 80 km/h when she sees a cyclist fall from his bike. Her reaction time is 1.8 seconds and her braking distance is 35 metres. What is her stopping distance? 80 × 1000 ÷ 60 ÷ 60 Step 1: Change 80 km/h into m/s = 22.2 m multiply by 1000 to go from km to m and then divide by 60 to go from hours to minutes and divide by 60 again to get to seconds reaction distance = 22.2 × 1.8
= 40 mStep 2: her reaction time is 1.8 seconds, so her reaction distance is 1.8 lots 22.2 40 m + 35 m Step 3: find her total stopping distance = reaction time + braking distance = 75 m in reply to: Stopping Distance, Distance, Speed and Time #18944The stopping distance (d) of a scooter travelling with a speed s metres per second is given by the formula d = ¼(s2 + s + 2). What is the stopping distance given a speed of 8 metres per second?
substitute s = 8 into the formula
s = ¼ × (82 + 8 + 2)
s = 18.5 m
in reply to: Stopping Distance, Distance, Speed and Time #18942Raine is driving at a speed of 80 km/h. It takes Raine two seconds to react to a dangerous situation before applying the brakes. The stopping distance (in m) is given by the formula: Stopping distance: $$d=\frac{5Vt}{18}+\frac{V^2}{170}$$. How far will Raine travel in her car after applying the brakes using this formula?
Substitute V = 80 and t = 2
$$d=\frac{5\times80\times2}{18}+\frac{80^2}{170}$$
= 82.0915
= 82 min reply to: Stopping Distance, Distance, Speed and Time #18939Sarah is travelling at 75 km/h and suddenly has to stop. Her reaction time is 4.2 seconds. She comes to a stop 116.19 m further down the road. Use the braking distance formula d = kv2 to find how much further, to the nearest metre, her stopping distance would be if she was travelling 12 km/h faster.
Stopping distance = braking distance + reaction time distance
Find the reaction time distance
75 km/h change to m/s by multiplying by 1000 to change km to m and dividing by 60 and then divide by 60 again to change from hours to minutes to seconds
speed in m/s = 75 × 1000 ÷ 60 ÷ 60
= 20.83 m/s
reaction time = 4.2 s
reaction time distance = 20.83 × 4.2= 87.5m
116.19 = braking distance + 87.5
braking distance = 116.19 -87.5
= 28.69 m
substitute this distance into the formula to find k
28.69 = k × 752
$$k=\frac{28.69}{5625}$$
k = 0.0051
∴ d = 0.0051v2
12 km/h faster = 75 + 12
= 87
find braking distance
d = 0.0051 × 872
d = 38.6m
find reaction distance
speed in m/s = 87 × 1000 ÷ 60 ÷ 60
= 24.16 m/s
reaction time distance = 24.16 × 4.2
= 101.5 m
stopping distance = 38.6 + 101.5
= 140.1 m
difference is 140.1 – 116.19
= 23.91 m
Find the value of H, correct to the nearest minute, in the formula $$BA{C_{Male}} = \frac{{10N – 7.5H}}{{6.8M}}$$ if:
a. $$BA{C_{Male}} =0.066$$, M = 60 and N = 5
b. $$BA{C_{Male}} =0.050$$, M = 70 and N = 7a. $$0.066=\frac{10\times5-7.5H}{6.8\times60}$$ substitute the values into the equation $$0.066=\frac{50-7.5H}{408}$$ multiply both sides by 408 26.928 = 50 – 7.5H subtract 50 from both sides -23.072 = -7.5H divide both sides by -7.5 H = 3.076 change to hours and minutes, using the degrees and minutes button H = 3 hours 5 minutes -
AuthorPosts
