Tanner will have to save $3,193.34 per month for 4 years to pay his tuition for Stanford.
Tanner's tuition over 4 years to get his Bachelor's degree all together will cost $185,280 (multiply 46,320×4). His parents will pay $32,000 (multiply 8,000×4) of his whole tuition. If we do the equation $185,280-$32,000 we get $153,280 which is what Tanner will have to pay. There are 48 months in 4 years. To find the answer you must solve the problem 48x= $153,280. To solve the problem divide both sides by 48. The answer to this equation is technically $3,193.3333333333 but for simplicity's sake we can round to $3,193.34. That is why the answer to this question is $3,193.34 per month of savings.
The first term is 138
The difference is 55
The iterative rule for the amount of money Mr Speas has after n weeks is
55/2 n² + 221/2 n
During the first week she has $138 in his bank account. At the end of each week she deposited $55 into her bank account.
The first term will be 138 .
The common difference is 55 because her bank always increase by $55 dollars every week. The sequence will be 138, 193, 248, 303, 358...…
The difference = 193 - 138 = 55.
The iterative rule for the amount of money Mr Speas has after n weeks can be represented below
n = number of weeks
a = first term = 138
d = common difference = 55
Using AP formula,
sₙ = n/2(2a + (n - 1)d)
sₙ = n/2 (2(138)+ (n - 1)55)
sₙ = n /2(276 + 55n -55)
sₙ = n /2(221 + 55n)
sₙ = 55/2 n² + 221/2 n
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Answer:
A)
Step-by-step explanation:
Starting point is at o and ending point at 2.
So, 2 in
Answer:
18/5
Step-by-step explanation:
<em>From my understanding, this is 1 whole 2 upon 10</em>
<u>Step 1: Convert mixed fraction into fraction</u>
<em>Multiply whole number by denominator: 1 x 10 = 10</em>
<em>Add numerator: 10 + 2 = 12</em>
<em>Write the result as the numerator and use the same denominator.</em>
12/10
<u>Step 2: Find the total tape used</u>
<em>From my understanding, each student used 12/10 of the tap</em>
<em>So 3 students used: 12/10 x 3 =</em> 18/5
Therefore, the total tape used for three students was 18/5.
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AB and AB’ are congruent
BC and B’C are congruent
All angles are congruent to the original angles
