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3 years ago

Line r has a slope of -5/2 line s has a slope of 2/5 are line r and line parallel or perpendicular

1 answer:

4 0

Vocab:
Parallel = same slope
Perpendicular = opposite sign and reciprocal slope

-5/2 opposite and reciprocal = 2/5

Solution: they are perpendicular

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Veronica is choosing between two health clubs after how many months will the total cost for each health club be the same

Assoli18 [71]

Question:

Veronica is choosing between two health clubs. after how many months will the total cost for each health club be the same? yoga studio a: membership: $24.00 monthly fee: 21.50. yoga studio b: membership: $41.00 monthly fee: $17.25

Answer:

It takes 4 years for the total cost of each club to become equal

Step-by-step explanation:

Given:

For yoga studio A:

membership: $24.00

monthly fee: 21.50.

For yoga studio B:

membership: $41.00

monthly fee: $17.25

To Find:

Number of months after which the total cost for each health club be the same = ?

Solution:

Let x be the number of months of membership, and y be equal the total cost.

For Yoga club A

y = 21.50 x + 24

For Yoga club B

y = 17.25 x + 41.00

we know that the prices, y , would be equal, we can set the two equations equal to each other.

21.50 x + 24 =17.25 x+ 41.00

Grouping the like terms,

21.50x - 17.25 x= 41.00 - 24

4.25x=17

x= $\frac{17}{4.25}$

x = 4

6 0

3 years ago

Which of the following equations is equivalent to log(y)= 3.994 ?

Kisachek [45]

Answer:

The correct answer is option D: $y = 10^{3.994}$

Step-by-step explanation:

Given:

log(y)= 3.994

Solution:

A logarithm base b of a positive number x satisfies the following definition:

$y =log_b(x) => b^y =x$

For $x > 0, b >0, b \neq 1$

Also if no base b is indicated, the base of the logarithm is assumed to be 10 .

Thus, in log(y)= 3.994 base b is not indicated. so its base is assumed to be 10

now

$log_{10}(y)= 3.994$

Then

$y = 10^{3.994}$

4 0

3 years ago

The table shows the height of a plant as it grows. Which equation in point slope form gives the plant’s height at any time

vodka [1.7K]

Answer:

Option A is correct.

$y-16=8(x-2)$ is the equation represent the point slope form gives the plant's height at any time.

Step-by-step explanation:

Point slope intercept form: For any two points $(x_1, y_1)$ and $(x_2, y_2)$ then,

the general form

$y-y_1=m(x-x_1)$ for linear equations; where m is the slope given by:

$m =\frac{y_2-y_1}{x_2-x_1}$

Consider any two points from the table;

let A= (2 , 16) and B =(4, 32)

First calculate the slope of the line AB:

$m =\frac{y_2-y_1}{x_2-x_1}=\frac{32-16}{4-2}=\frac{16}{2}$ = 8

Therefore, slope of the line m = 8

Then,

the equation of line is:

$y-y_1=m(x-x_1)$

Substitute the value of m=8 and (2, 16) above we get;

$y-16=8(x-2)$

Therefore, the equation in point slope form which gives the plant's height at any time is; $y-16=8(x-2)$ , where x is the time(months) and y is the plant height (cm)

5 0

3 years ago

Read 2 more answers

Please give me the answer

Tcecarenko [31]

Answer:

7 units

Step-by-step explanation:

hope I helped today

8 0

2 years ago

Read 2 more answers

Please answer correctly !!!!!!!!! Will<br> Mark brainliest !!!!!!!!!!!!!

il63 [147K]

Answer:

$x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

Step-by-step explanation:

$w=-5\left(x-8\right)\left(x+4\right)\\\mathrm{Expand\:}-5\left(x-8\right)\left(x+4\right):\quad -5x^2+20x+160\\w=-5x^2+20x+160\\Switch\:sides\\-5x^2+20x+160=w\\\mathrm{Subtract\:}w\mathrm{\:from\:both\:sides}\\-5x^2+20x+160-w=w-w\\Simplify\\-5x^2+20x+160-w=0\\Solve\:with\:the\:quadratic\:formula\\\mathrm{Quadratic\:Equation\:Formula:}\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-5,\:b=20,\:c=160-w:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}\\x=\frac{-20+\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20+\sqrt{-20w+3600}}{10}\\x=\frac{-20-\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20-\sqrt{-20w+3600}}{10}\\The\:solutions\:to\:the\:quadratic\:equation\:are\\x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

6 0

3 years ago