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umka21 [38]
4 years ago
8

Line m passes through (4, 7) and (8, 2). Line n is parallel to line m. What is the slope

Mathematics
1 answer:
Harlamova29_29 [7]4 years ago
4 0
X= 4 and Y=-5
or
X=-4 and Y=5
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Can someone help me with answer C? its the last one i need
Dmitry [639]

Using the monthly payment formula, it is found that her down payment should be of $1,419.

<h3>What is the monthly payment formula?</h3>

It is given by:

A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}

In which:

  • P is the initial amount.
  • r is the interest rate.
  • n is the number of payments.

For this problem, the parameters are:

A = 250, r = 0.072, n = 72.

Hence:

r/12 = 0.072/12 = 0.006.

We solve for P to find the total amount of the monthly payments, hence:

A = P\frac{\frac{r}{12}\left(1 + \frac{r}{12}\right)^n}{\left(1 + \frac{r}{12}\right)^n - 1}

P\frac{0.006(1.006)^{72}}{(1.006)^{72}-1} = 250

0.0171452057P = 250

P = 250/0.0171452057

P = $14,581.

The total payment is of $16,000, hence her down payment should be of:

16000 - 14581 = $1,419.

More can be learned about the monthly payment formula at brainly.com/question/26476748

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4 0
2 years ago
Genuinely confused rn
frosja888 [35]

Using it's concept, it is found that the probability that he winds up wearing the white shirt and tan pants is of \frac{1}{8}.

<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

In this problem:

  • For the shirt, there is 4 outcomes, hence the probability of the white shirt is 1/4.
  • For the pair of pants, there are 2 outcomes, blue or tan, hence the probability of tan pants is 1/2.

Since the shirt and the pants are independent, the probability is given by:

p = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}.

More can be learned about probabilities at brainly.com/question/14398287

#SPJ1

8 0
2 years ago
Okay will award brainliest and put alot of points so pls help I'm desparate!!!
Bad White [126]

Answer:

23 is 31/100

Step-by-step explanation:

not sure about the rest sorry :(

7 0
4 years ago
Read 2 more answers
The high for a winter day in Sitka,
m_a_m_a [10]

Answer:

The maximum temperature will be -10°C, then if T represents the temperature, we can write this as:

T ≤ -10°C

And the minimum temperature will be -25°C, then we must have that:

T ≥ -25°C

if we use both conditions, we will have:

-25°C ≤ T ≤ -10°C

We can write this range:

[-25°C, -10°C]

Where the [] symbols mean that the extremes are possible temperatures.

The length of the range will be:

-10°C - 25°C = 15°C.

3 0
3 years ago
5^(-x)+7=2x+4 This was on plato
Setler79 [48]

Answer:

Below

I hope its not too complicated

x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

Step-by-step explanation:

5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}

Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}

\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x

\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

3 0
3 years ago
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