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Vikki [24]
3 years ago
9

Javier works in a photo lab and needs to make a reduced copy of a photograph that is 8 inches long. The customer wants the copy

to have a length of 4.5 inches. Each time he presses the reduce button on the copier, the copy is reduced by 12%. Javier and his boss had an argument, because Javier claimed that pressing the button 5 times would be enough to fulfill the customer's request. His boss said that pressing the button 5 times would result in a length less than the desired 4.5 inches. Which of the following conclusions can be made?
Mathematics
1 answer:
Volgvan3 years ago
4 0
So, there's total 8 inches and the customer wants to make 4.5 inches long.
Each time he press the reduce button is 12%. He pressed 5 times, so.

12 x 5 = 60%
60% of the 8 = is 4.8 inches

So the conclusion is that his boss is wrong.
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Calvin deposits $400 in a savings account that accrues 5% interest compounded monthly. after c years, calvin has $658.80. makayl
vodomira [7]
You will need a special formula to compute this.
Years = log (total/principal) / [n * log * (1 + rate / n)]

Part A) Calvin $400 5% monthly 658.80 Time = ?
Monthly compounding "n" = 12
Years = log(658.80/400) / [12 * log(1+ (.05/n))
Years = log ( <span> <span> <span> 1.647 </span> </span> </span> ) / (12 * log ( <span><span><span>1.0041666667 </span> </span> </span> )
Years = 0.21669359917 / 12 * 0.0018058008777
Years = <span> <span> </span></span><span><span>0.21669359917 / 0.0216696105 </span>
</span>Years = <span><span>9.999884362 </span>

</span>
Part B) Makayla 300 6% quarterly  613.04Time=?
Quarterly compounding
n = 4
Years = log (total/principal) / [n * log * (1 + rate / n)]
Years = log (613.04/300) / [4 * log (1 + .06/4)]
Years = log ( <span> <span> <span> 2.0434666667 </span> </span> </span> ) / 4 * log (1.015)
Years = 0.31036755784 / 4 * 0.0064660422492
Years = 0.31036755784 / <span> <span> <span> 0.025864169 </span> </span> </span>
Years = <span> <span> <span> 11.9999044949 </span> </span> </span>

So, the difference is roughly 3 years.



3 0
3 years ago
Read 2 more answers
Can you tell me the definition of the highlighted word please in your own words (Thank you)
sattari [20]

Answer:

congruent means similar

Step-by-step explanation:

why didnt you just search it up? lol

8 0
3 years ago
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What are the coordinates of the midpoint of EF?
BabaBlast [244]
The answer would be d. (1,3.5)
you would use the midpoint formula that is attached below. 

7 0
3 years ago
A cashier has a total of 30 bills consisting of ones, fives and twenties. the number of twenties is 5 less than the number of on
Varvara68 [4.7K]

Let us say that:

a = ones

b = fives

c = twenties

 

So that the total money is:

1 * a + 5 * b + 20 * c = 229

=> a + 5b + 20c = 229                                       --> eqtn 1

 

We are also given that:

c = a – 5                                                                --> eqtn 2

a + b + c = 30                                                      --> eqtn 3

 

Rewriting eqtn 3 in terms of b:

b = 30 – a – c

Plugging in eqtn 2 into this:

b = 30 – a – (a – 5)

b = 35 – 2a                                                           --> eqtn 4

 

Plugging in eqtn 2 and 4 into eqtn 1:

a + 5(35 – 2a) + 20(a – 5) = 229

a + 175 – 10a + 20a – 100 = 229

11a = 154

a = 14

 

So,

b = 35 – 2a = 7

c = a – 5 = 9

 

Therefore there are 14 ones, 7 fives, and 9 twenties.

7 0
4 years ago
The vector (a) is a multiple of the vector (2i +3j) and (b) is a multiple of (2i+5j) The sum (a+b) is a multiple of the vector (
kow [346]

Answer:

\|a\| = 5\sqrt{13}.

\|b\| = 3\sqrt{29}.

Step-by-step explanation:

Let m,n, and k be scalars such that:

\displaystyle a = m\, (2\, \vec{i} + 3\, \vec{j}) = m\, \begin{bmatrix}2 \\ 3\end{bmatrix}.

\displaystyle b = n\, (2\, \vec{i} + 5\, \vec{j}) = n\, \begin{bmatrix}2 \\ 5\end{bmatrix}.

\displaystyle (a + b) = k\, (8\, \vec{i} + 15\, \vec{j}) = k\, \begin{bmatrix}8 \\ 15\end{bmatrix}.

The question states that \| a + b \| = 34. In other words:

k\, \sqrt{8^{2} + 15^{2}} = 34.

k^{2} \, (8^{2} + 15^{2}) = 34^{2}.

289\, k^{2} = 34^{2}.

Make use of the fact that 289 = 17^{2} whereas 34 = 2 \times 17.

\begin{aligned}17^{2}\, k^{2} &= 34^{2}\\ &= (2 \times 17)^{2} \\ &= 2^{2} \times 17^{2} \end{aligned}.

k^{2} = 2^{2}.

The question also states that the scalar multiple here is positive. Hence, k = 2.

Therefore:

\begin{aligned} (a + b) &= k\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 2\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 16\, \vec{i} + 30\, \vec{j}\\ &= \begin{bmatrix}16 \\ 30 \end{bmatrix}\end{aligned}.

(a + b) could also be expressed in terms of m and n:

\begin{aligned} a + b &= m\, (2\, \vec{i} + 3\, \vec{j}) + n\, (2\, \vec{i} + 5\, \vec{j}) \\ &= (2\, m + 2\, n) \, \vec{i} + (3\, m + 5\, n) \, \vec{j} \end{aligned}.

\begin{aligned} a + b &= m\, \begin{bmatrix}2\\ 3 \end{bmatrix} + n\, \begin{bmatrix} 2\\ 5 \end{bmatrix} \\ &= \begin{bmatrix}2\, m + 2\, n \\ 3\, m + 5\, n\end{bmatrix}\end{aligned}.

Equate the two expressions and solve for m and n:

\begin{cases}2\, m + 2\, n = 16 \\ 3\, m + 5\, n = 30\end{cases}.

\begin{cases}m = 5 \\ n = 3\end{cases}.

Hence:

\begin{aligned} \| a \| &= \| m\, (2\, \vec{i} + 3\, \vec{j})\| \\ &= m\, \| (2\, \vec{i} + 3\, \vec{j}) \| \\ &= 5\, \sqrt{2^{2} + 3^{2}} = 5 \sqrt{13}\end{aligned}.

\begin{aligned} \| b \| &= \| n\, (2\, \vec{i} + 5\, \vec{j})\| \\ &= n\, \| (2\, \vec{i} + 5\, \vec{j}) \| \\ &= 3\, \sqrt{2^{2} + 5^{2}} = 3 \sqrt{29}\end{aligned}.

6 0
3 years ago
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