Solve for y.<br><br> –4(y − 15) + –6 =

A fitness center has two membership plans. One has a low dollar sign up fee of 15 dollars and cost 38 dollars per month to join.

victus00 [196]

Let x be the number of months.
The first plan is 15 dollars sign up fee and 38 dollars per month. So the equation is 38x + 15.
The second plan is 78 dollars as sign up fee and 31 dollars per month. So the equation is 31x + 78.

We need find when x has the same value in both equations, so we do their equality:
38x + 15 = 31x + 78
Let's subtract 15 from both sides
38x + 15 = 31x + 78
38x + 15 - 15 = 31x + 78 - 15
38x = 31x + 63

Now let's subtract 31x from both sides to have the variables on a side and the numbers on side:

38x = 31x + 63
38x - 31x = 31x - 31x + 63
7x = 63

Divide both sides by 7 to have the variable x on a side and its value on the other:
(7x)/7 = 63/7
x = 9

So at month 9, the 2 plans will cost the same.

Let's check our answers, and let y be the cost:
y = 38x + 15 = 38*9 + 15 = 357
y = 31x + 78 = 31*9 + 78 = 357
Our answer has been approved.

Hope this helps! :D

8 0

3 years ago

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lilavasa [31]

The count of the equilateral triangle is an illustration of areas

There are 150 small equilateral triangles in the regular hexagon

<h3>How to determine the number of equilateral triangle </h3>

The side length of the hexagon is given as:

L = 5

The area of the hexagon is calculated as:

$A = \frac{3\sqrt 3}{2}L^2$

This gives

$A = \frac{3\sqrt 3}{2}* 5^2$

$A = \frac{75\sqrt 3}{2}$

The side length of the equilateral triangle is

l = 1

The area of the triangle is calculated as:

$a = \frac{\sqrt 3}{4}l^2$

So, we have:

$a = \frac{\sqrt 3}{4}*1^2$

$a = \frac{\sqrt 3}{4}$

The number of equilateral triangles in the regular hexagon is then calculated as:

$n = \frac Aa$

This gives

$n = \frac{75\sqrt 3}{2} \div \frac{\sqrt 3}4$

So, we have:

$n = \frac{75}{2} \div \frac{1}4$

Rewrite as:

$n = \frac{75}{2} *\frac{4}1$

$n = 150$

Hence, there are 150 small equilateral triangles in the regular hexagon

Read more about areas at:

brainly.com/question/24487155

4 0

3 years ago

At 106°F, a certain insect chirps at a rate of 67 times per minute, and at 108°F, they chirp 83 times per minute. Write an equat

Step2247 [10]

Answer:

$y=8x-781$

Here, $x$ represents temperature and $y$ denotes rate of chirping per minute.

Step-by-step explanation:

Let $x$ represents temperature and $y$ denotes rate of chirping per minute.

At $106$ °F, a certain insect chirps at a rate of $67$ times per minute.

Take $(x_1,y_1)=(106,67)$

At $108$ °F, they chirp $83$ times per minute.

Take $(x_2,y_2)=(108,83)$

Slope intercept form:

$y-y_1=(\frac{y_2-y_1}{x_2-x_1})(x-x_1)$

$y-67=(\frac{83-67}{108-106})(x-106)\\\\y-67=(\frac{83-67}{108-106})(x-106)\\\\y-67=(\frac{16}{2})(x-106)\\\\y-67=8(x-106)\\y-67=8x-848\\y=8x+67-848\\y=8x-781$

3 0

3 years ago

Please Help

e-lub [12.9K]

The general formula for the equation of the straight line is:
y−b=m(x−a) where
m is the slope
a is the x coordinate of the given point
b is the y coordinate of the given point

The slope is given as 2
x coordinate in the given point is 5
y coordinate of the given point is 9

Substitute with these values in the mentioned equation as follows:
y−9=2(x-5)

4 0

3 years ago

Read 2 more answers

Why is everyone ignoring the questions I ask?

Elena L [17]

Answer:

o-

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Solve for y. –4(y − 15) + –6 = –10 What does y equal