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Vsevolod [243]
3 years ago
5

Can you do c and d pla

Mathematics
1 answer:
Anna71 [15]3 years ago
7 0

Answer:

c) y = - 2 x +3

d)  y = 1/2  x + 1

Step-by-step explanation:

For c) Look for the exression that contains "3" as the y-intercept (the "b" in the form y = m x + b)

The only equation is y = - 2 x +3

For d) try in all equations if replacing x with 2, gives you an answer "2" for y as well.

The only equation that does such is   y = 1/2  x + 1

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A vehicle is purchased for $18,000, with a down payment of $6,098. The balance in financed for three years at an annual rate of
mestny [16]
The Present value of an annuity is given by PV = P(1 - (1 + r/t)^-nt)/(r/t)
where: P is the monthly payment, r is the annual rate = 7% = 0.07, t is the number of periods in one year = 12 and n is the number of years = 3.

18,000 - 6,098 = P(1 - (1 + 0.07/12)^-(3 x 12)) / (0.07/12)
11,902 = P(1 - (1 + 0.07/12)^-36) / (0.07/12)
P = 0.07(11,902) / 12(1 - (1 + 0.07/12)^-36) = 367.50

Therefore, monthly payment = $367.50
7 0
3 years ago
If you answer this correctly i will cash app you 20 dollars ​
Law Incorporation [45]
225pi because I know ..
6 0
3 years ago
Estimate the following quantities without using a calculator. Then find a more precise result, using a calculator if necessary.
zubka84 [21]

Explanation:

a. It is so easy to double a number that no approximation is necessary.

  31,000 × 200 = 3.1×10⁴ × 2×10² = 6.2×10⁶ exactly

__

b. 6.2×10³ × 5.2×10⁶ ≈ 6×5×10⁹ = 3×10¹⁰ approximately

The approximation can be refined a bit by taking the ".2" into account:

  6.2×10³ × 5.2×10⁶ ≈ (6×5 + .2×(6+5))×10⁹

  = 32.2×10⁹ = 3.22×10¹⁰ approximately

Actual product: 3.224×10¹⁰.

For most purposes, the approximation is an adequate approximation, as it it within 10% of the actual value.

__

c. 9×10⁶ -2.3×10⁴ ≈ 9×10⁶ approximately

A better approximation is to actually subtract an approximation of the smaller number:

  9×10⁶ -2.3×10⁴ ≈ (9 - 0.02)×10⁶ ≈ 8.98×10⁶ approximately

The actual value uses all of the digits of the smaller number:

  9×10⁶ -2.3×10⁴ ≈ (9 - 0.023)×10⁶ = 8.977×10⁶ exactly

As in part B, either approximation is adequate for most purposes, as the difference from the actual value is less than .3%.

_____

The accuracy required of an approximation, hence the work you expend improving accuracy, should depend on the need in the final application of the number. Often, approximations are used for budget or resource planning purposes where some "slop" is allowed or even expected.

They can also be used in engineering applications, where the error needs to be on the side of more safety (rather than less).

7 0
3 years ago
What is the product of 1/2 times 2/3 times 3/4 .... times 1999/2000?
Phantasy [73]
1/2 * 2/3 *3/4 * ... * 1999/2000
Observe, every numerator cancels with denominator
that leaves 1/2000
3 0
3 years ago
3.
Lerok [7]

3. Here we use the formula

A = P ( 1 + \frac{r}{n} )^ {nt}

And we have the following values given

P = $23400, r = 3%=0.03,n =2, t=10 years

So we will get

A = 23400(1+ \frac{0.03}{2} )^{2*10}
\\
A = 23400( \frac{2.03}{2} )^{20} = $31516.52

Question 4.

In this question , we have

P = $2310, R = 3.5% = 0.035 ,
\\
Number \ of \ days \ from \ april \12 to July \ 5 = 30+31+23 = 84 days

A = 2310(1+ \frac{0.035}{365})^{84}  = $2328.68

Interest is the difference of amount and principal, that is

I = 2328.68-2310 = $18.68

5 0
3 years ago
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