Answer:
1. <2, <7
2. 58
3. 78
4. a and b
h5. <2 and <7 ( i think this is the same question as the first one)
Step-by-step explanation:
2. 2x + 7 = 123 (you have to make them equal to each other because they are consecutive interior angles)
2x + 7 = 123, subtract 7
2x = 116, divide 2 for both sides to make x by itself
x = 58
3. 78 because of consecutive interior angles
<u>Hint </u><u>:</u><u>-</u>
- Break the given sequence into two parts .
- Notice the terms at gap of one term beginning from the first term .They are like
. Next term is obtained by multiplying half to the previous term . - Notice the terms beginning from 2nd term ,
. Next term is obtained by adding 3 to the previous term .
<u>Solution</u><u> </u><u>:</u><u>-</u><u> </u>
We need to find out the sum of 50 terms of the given sequence . After splitting the given sequence ,
.
We can see that this is in <u>Geometric</u><u> </u><u>Progression </u> where 1/2 is the common ratio . Calculating the sum of 25 terms , we have ,
![\implies S_1 = a\dfrac{1-r^n}{1-r} \\\\\implies S_1 = \dfrac{1}{2}\left[ \dfrac{1-\bigg(\dfrac{1}{2}\bigg)^{25}}{1-\dfrac{1}{2}}\right]](https://tex.z-dn.net/?f=%5Cimplies%20S_1%20%3D%20a%5Cdfrac%7B1-r%5En%7D%7B1-r%7D%20%5C%5C%5C%5C%5Cimplies%20S_1%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Cleft%5B%20%5Cdfrac%7B1-%5Cbigg%28%5Cdfrac%7B1%7D%7B2%7D%5Cbigg%29%5E%7B25%7D%7D%7B1-%5Cdfrac%7B1%7D%7B2%7D%7D%5Cright%5D%20)
Notice the term 
will be too small , so we can neglect it and take its approximation as 0 .
![\implies S_1\approx \cancel{ \dfrac{1}{2} } \left[ \dfrac{1-0}{\cancel{\dfrac{1}{2} }}\right]](https://tex.z-dn.net/?f=%5Cimplies%20S_1%5Capprox%20%5Ccancel%7B%20%5Cdfrac%7B1%7D%7B2%7D%20%7D%20%5Cleft%5B%20%5Cdfrac%7B1-0%7D%7B%5Ccancel%7B%5Cdfrac%7B1%7D%7B2%7D%20%7D%7D%5Cright%5D%20)
Now the second sequence is in Arithmetic Progression , with common difference = 3 .
![\implies S_2=\dfrac{n}{2}[2a + (n-1)d]](https://tex.z-dn.net/?f=%5Cimplies%20S_2%3D%5Cdfrac%7Bn%7D%7B2%7D%5B2a%20%2B%20%28n-1%29d%5D%20)
Substitute ,
![\implies S_2=\dfrac{25}{2}[2(4) + (25-1)3] =\boxed{ 908}](https://tex.z-dn.net/?f=%5Cimplies%20S_2%3D%5Cdfrac%7B25%7D%7B2%7D%5B2%284%29%20%2B%20%2825-1%293%5D%20%3D%5Cboxed%7B%20908%7D%20)
Hence sum = 908 + 1 = 909
Answer:
Bc = √63 ft
Step-by-step explanation:
Here, we want to get the length of BC
As we can see, there is a right angled triangle ABC, with 12ft being the hypotenuse and 9 ft the other side
So we want to get the third leg which is BC
we can use the Pythagoras’ theorem here
And that states that the square of the hypotenuse equals the sum of the squares of the two other sides
let the missing length be x
12^2 = x^2 + 9^2
144 = x^2 + 81
x^2 = 144-81
x^2 = 63
x = √63 ft
To solve this problem, let us first assign variables. Let
us say that:
A = runner
B = cyclist
d = distance
v = velocity
time = t
The time in which the cyclist overtakes the runner is the
time wherein the distance of the two is the same, that is:
dA = dB
We know that the formula for calculating distance is:
d = v t
therefore,
vA tA = vB tB
Further, we know that tA = tB + 2, therefore:
vA (tB + 2) = vB tB
4 (tB + 2) = 14 tB
4 tB + 8 = 14 tB
10 tB = 8
tB = 0.8 hours = 48 min
Therefore the cyclist overtakes the runner after 0.8
hours or 48 minutes.
