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3 years ago

Y+1=2/5(x+3) in standard form?

Mathematics

1 answer:

Rzqust [24]3 years ago

6 0

For this case we have that by definition, the equation of a line in the standard form is given by:

$ax + by = c$

According to the statement we have the following equation of the line:

$y + 1 = \frac {2} {5} (x + 3)$

We manipulate algebraically to write in the standard form. To do this we follow the steps below:

We apply distributive property on the right side of the equation: $y + 1 = \frac {2} {5} x + \frac {6} {5}$

We subtract $\frac {6} {5}$ on both sides of the equation:

$y + 1- \frac {6} {5} = \frac {2} {5} x\\y- \frac {1} {5} = \frac {2} {5} x$

We multiply by 5 on both sides of the equation:

$5y-1 = 2x$

We subtract 5y on both sides of the equation:

$2x-5y = -1$

Finally, the equation in its standard form is:

$2x-5y = -1$

Answer:

$2x-5y = -1$

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A package of 6 pairs of insulated gloves costs 38.34. What is the unit price of the pairs of gloves?

devlian [24]

Answer:

6.39

Step-by-step explanation:

divide the total cost by the total amount of the item for the unit price:

38.34/6=6.39

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3 years ago

EVALUATE THE WORD PROBLEMS

andrezito [222]

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3 years ago

logan trades 50 pennies for nickels he gets a nickel for each group of 5 pennies what equation can be used to find how many nick

V125BC [204]

The equation that can be used to find how many nickels logan uses is: 50÷5.

<h3>Equation to find the number of nickels</h3>

Given:

Total pennies=50

Logan got a nickel of 5 pennies=1

Hence:

Number of nickels=1÷5×50

Number of nickels=50÷5

Number of nickels=10 nickels

Therefore the equation that can be used to find how many nickels logan uses is: 50÷5.

Learn more about Equation to find the number of nickels here:brainly.com/question/10704332

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2 years ago

A valid license plate in Xanadu consists of two capital letters followed by three digits. How many valid license plates are poss

tatiyna

The valid license plates are possible = 676,000 possibilities

<h3>What is permutation and combination?</h3>

In mathematics, there are two alternative methods for dividing up a collection of components into subsets: combination and permutation. The subset's components can be arranged in any sequence when used together. The subset's components are given in a permutation in a certain order.

<h3>According to the given information:</h3>

There is no prohibition on using the same letters or numbers more than once, therefore all 26 letters of the alphabet and all 10 numerals are permissible choices.

There are 26 options if the initial answer is A:

AA, AB, AC,AD,AE ...................................... AW, AX, AY, AZ.

There are 26 options if the initial answer is B:

BA, BB, BC, BD, BE .........................................BW, BX,BY,BZ

So,

The first letter can be one of 26 options, and the second letter can be one of 26 options. There are a variety of 2 letter combinations, including:

26 x 26 = 676

For the three digits, the same holds true.

The first, second, and third options each have ten options available:

10 × 10 × 10 = 1000

Thus, there are several options for a license plate with two letters and three digits:

26 x 26 x 10 × 10 × 10 = 676,000 possibilities.

The valid license plates are possible = 676,000 possibilities.

To know more about permutation and combination visit:

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8 0

1 year ago

6. If the net investment function is given by

Pachacha [2.7K]

The capital formation of the investment function over a given period is the

accumulated capital for the period.

(a) The capital formation from the end of the second year to the end of the fifth year is approximately <u>298.87</u>.

(b) The number of years before the capital stock exceeds $100,000 is approximately <u>46.15 years</u>.

Reasons:

(a) The given investment function is presented as follows;

$I(t) = 100 \cdot e^{0.1 \cdot t}$

(a) The capital formation is given as follows;

$\displaystyle Capital = \int\limits {100 \cdot e^{0.1 \cdot t}} \, dt =1000 \cdot e^{0.1 \cdot t}} + C$

From the end of the second year to the end of the fifth year, we have;

The end of the second year can be taken as the beginning of the third year.

Therefore, for the three years; Year 3, year 4, and year 5, we have;

$\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87$

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87

(b) When the capital stock exceeds $100,000, we have;

$\displaystyle \mathbf{\left[1000 \cdot e^{0.1 \cdot t}} + C \right]^t_0} = 100,000$

Which gives;

$\displaystyle 1000 \cdot e^{0.1 \cdot t}} - 1000 = 100,000$

$\displaystyle \mathbf{1000 \cdot e^{0.1 \cdot t}}} = 100,000 + 1000 = 101,000$

$\displaystyle e^{0.1 \cdot t}} = 101$

$\displaystyle t = \frac{ln(101)}{0.1} \approx 46.15$

The number of years before the capital stock exceeds $100,000 ≈ <u>46.15 years</u>.

Learn more investment function here:

brainly.com/question/25300925

6 0

3 years ago