Jordan is starting to save $150 to buy a new cell phone. In December, he saved $5. In January, he plans twice as much as he did
1 answer:
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Answer:
April
Step-by-step explanation:
The sequence of amounts saved is a geometric sequence. The sum of those amounts is given by the formula ...
Sn = a1·(r^n -1)/(r -1)
where a1 is the amount saved in month 1, r is the common ratio, and n is the number of months. We want to find n such that ...
150 ≤ 5(2^n -1)/(2 -1)
30 ≤ 2^n -1
31 ≤ 2^n
We can solve this using logarithms, or we can make use of our knowledge of powers of 2: 2^5 = 32. That is, n = 5 will be the first month when Jordan's total savings exceeds $150.
The fifth month will be April. Jordan's total savings will be enough in April.
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The attached shows a spreadsheet that computes Jordan's amount of savings each month, and his running total. Once the formulas are written, they can be copied down as far as required to see where the total savings is enough.
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Answer:
∠BKM= ∠ABK
Therefore AB ║KM (∵ ∠BKM= ∠ABK and lies between AB and KM and BK is the transversal line)
m∠MBK ≅ m∠BKM (Angles opposite to equal side of ΔBMK are equal)
Step-by-step explanation:
Given: BK is an angle bisector of Δ ABC. and line KM intersect BC such that, BM = MK
TO prove: KM ║AB
Now, As given in figure 1,
In Δ ABC, ∠ABK = ∠KBC (∵ BK is angle bisector)
Now in Δ BMK, ∠MBK = ∠BKM (∵ BM = MK and angles opposite to equal sides of a triangle are equal.)
Now ∵ ∠MBK = ∠BKM
and ∠ABK = ∠KBM
∴ ∠BKM= ∠ABK
Therefore AB ║KM (∵ ∠BKM= ∠ABK and BK is the transversal line)
Hence proved.
