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Marizza181 [45]
3 years ago
10

Indicate the equation of the given line in standard form.

Mathematics
1 answer:
s2008m [1.1K]3 years ago
6 0

The line through (a,b) with slope m is

y-b = m(x-a)

Here that's

y + 1 = (3/4) (x - 2)

Let's multiply by 4 to clear the fractions and distribute the constants,

4y + 4 = 3x - 6

Rearranging into standard form,

3x - 4y = 10

Check.  The slope is 3/4, good. 3(2) - 4(-1) = 6+4 = 10 good

Answer: 3x - 4y = 10

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How to find compound rate from given values, 10 years after, and initial
dybincka [34]

The formula that calculates the compound rate from the given values is r = n(-1 + \sqrt[10n]{\frac{P + I}{P}})

<h3>How to determine the compound interest rate?</h3>

The compound interest formula is:

I = P(1 + \frac rn)^{nt} - P

Where:

  • P represents the principal amount
  • r represents the compound interest rate
  • n represents the number of times the interest is compounded
  • t represents the time in years
  • I represents the interest

We start by adding P to both sides

P + I = P(1 + \frac rn)^{nt}

Divide through by P

\frac{P + I}{P} = (1 + \frac rn)^{nt}

Take the nt-th root of both sides

\sqrt[nt]{\frac{P + I}{P}} = 1 + \frac rn

Subtract 1 from both sides

-1 + \sqrt[nt]{\frac{P + I}{P}} = \frac rn

Multiply through by n

r = n(-1 + \sqrt[nt]{\frac{P + I}{P}})

In this case, t = 10

So, we have:

r = n(-1 + \sqrt[10n]{\frac{P + I}{P}})

Hence, the formula that calculates the compound rate is r = n(-1 + \sqrt[10n]{\frac{P + I}{P}})

Read more about compound interest at:

brainly.com/question/13155407

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3 0
2 years ago
$10 000 is borrowed at 10.5% per annum interest. How much interest is owed at the end of the first 10 months?
Leviafan [203]

Answer:

$875 is the interest owed at the end of the first 10 months

Step-by-step explanation:

In this question, we are to calculate how much interest is owed in the first ten months.

There are several routes to this but the one i will like to use is as follows.

we shall calculate the amount owed for a year, afterwards we did use this amount by 12 since there are 12 months in a year. Now we multiply whatever answer we have by 10 to get the amount owed in 10 months.

To calculate the interest, we use the simple interest formula.

Mathematically;

I = PRT/100

where P is the amount borrowed which is $10,000

R is the interest rate which is 10.5% per annum

T is the time which we agreed we would be using 1 year first.

I = (10,000 * 10.5 * 1)/100 = $1050

This is the amount in 12 months. In a single month, this will be 1050/12 = $87.5

This amount in 10 months will be 87.5 * 10 = $875

8 0
3 years ago
Can someone help me ASAP!!!
Nostrana [21]

9514 1404 393

Answer:

  f(x) = {-x for x < 2; 2x-3 for x ≥ 2}

Step-by-step explanation:

The blue line extending to the left has a slope of -1 and a y-intercept of 0. Its equation is y = -x. That part of the definition of f(x) is applicable for values of x less than 2. (The open dot at (2, -2) tells you that point is not included.)

The red line extending to the right has a slope = rise/run = 2/1 = 2. The y-intercept can be found by extending the line to the y-axis, or from the computation ...

  b = y -mx

  b = 1 -2(2) = -3 . . . . . for the point (x, y) = (2, 1)

Then the slope-intercept equation for the red line is ...

  y = 2x -3

That part of the definition of f(x) is applicable for values of x ≥ 2. The solid dot tells you the point (2, 1) is included.

Putting these parts together, we get ...

  f(x)=\left\{\begin{array}{cc}-x&\text{if $x

8 0
3 years ago
An art teacher has a total of 5/8 pound of clay.The teacher puts 1/32 pound of clay at each work station.How many total work sta
defon

Answer: Total work stations = 20

Step-by-step explanation:

Total clay = \dfrac58 pound

Clay at each work station =  \dfrac1{32} pound

Total work stations = (Total clay ) ÷ (Clay at each work station )

=\dfrac58\div\dfrac{1}{32}\\\\=\dfrac58\times32\\\\= 5\times4\\\\= 20

Hence, Total work stations = 20

7 0
3 years ago
What is 3.25555 as a fraction?
Yuliya22 [10]
3.25555 as a fraction<span> equals </span><span>325555/100000</span>
7 0
3 years ago
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