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3 years ago

If y varies directly with x and y = 25 when x = 5, what is the value of k? how does this work when y = kx?

Mathematics

2 answers:

NISA [10]3 years ago

7 0

<h2>Answer:</h2>

The value of k when x=5 and y=25 is:

k=5

<h2>Step-by-step explanation:</h2>

Direct variation--

Two variables are said to be in direct variation if one variable could be written as a constant multiple of the other variable.

i.e. x and y are said to be in direct variation.

if there exist a constant k such that:

$y=kx$

or if x is non-zero then we have:

$k=\dfrac{y}{x}$

Here we are given x=5 and y=25

This means that the value of k is:

$k=\dfrac{25}{5}\\\\i.e.\\\\k=5$

Savatey [412]3 years ago

3 0

If y = 25 and x = 5

25 = 5k

k = 5

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Help me pls pls pls

cestrela7 [59]

Answer:

I guess the answer is A.

$\frac{1}{5}$

4 0

3 years ago

Use calculus to find the area of the triangle with the vertices (0, 5), (2, -2), and (5, 1).

guajiro [1.7K]

The area of the triangle with the vertices (0, 5), (2, -2), and (5, 1) by using the calculus is 21 square unit.

We need to find the equation among all possible pairs and then integrate the equations from one co-ordinate to another co-ordinate

Equation of line passing through (0,5) and (2,-2) is

y-5 = [(-2-5)/(2-0)](x-0)

=>y-5 = (-7) /2x

=>y= (-7/2x)+5 -------(eq1)

Equation of line passing through (0,5) and (5,1) is

y-5 = [(1-5) / (5-0)](x-0)

=>y-5 = (-4/5)x

=>y = (-4/5)x+5-------(eq2)

Equation of line passing from (2,-2) and (5,1) is

y-(-2) = [[1 - (-2)] / (5-2)] / (x-2)

=>y+2=(3/3)(x-2)

=>y=x-4--------(eq3)

Now, we use definite integration to find the area between the different equation of line.

So, area enclosed between the equations is given by the

area = $\int\limits^5_2$ [(-4/5)x+5 - (-7/2)x + 5)dx + $\int\limits^5_1$ [(-4/5)x+5 -(x-4)]dx

=>area= $\int\limits^5_2$ (7/2-4/5)x dx + $\int\limits^5_1$ ((-9/5)x+9)dx

Using properties of integration, $\int\limits x\, dx=x^{2}/2$

=>area= $\int\limits^5_2$ (27/10)x dx + $\int\limits^5_1$ (-9/5)x+9)dx

=>area=([27/10)×[5² - 2²])/2 + [ (-9/5)×(5²-1²) ]/2 +9×(5-1)

=>area=(27/20)×(25-4) + (-9/5)×24+9×4

=>area = (27×21)/20 + (-216)/5+ 36

=>area=(567/20) - (216/5) + 36

=>area= [(567-261×4)+(36×20)]/20

=>area=[(567-864)+720]/20

=>area=423/20

=>area=21 square unit.

Hence, area of triangle is 21 square unit.

To know more about area of triangle, visit here:

brainly.com/question/19305981

#SPJ4

4 0

1 year ago

Determine if this relationship forms a direct variation. Verify your answer.

leva [86]

we know that

A relationship between two variables, x, and y, represent a direct variation if it can be expressed in the form y/x=k or y=kx

we have

a) for x=1 y=0.50--------> y/x=0.50/1------> 0.50

b) for x=2 y=1--------> y/x=1/2-------> 0.50

c) for x=3 y=1.50--------> y/x=1.50/3-----> 0.50

d) for x=5 y=2.50--------> y/x=2.50/5----> 0.50

the value of k is equal to 0.50

<span>the relationship forms a direct variation. </span>

the equation is

y=0.50*x

Verify

for x=1

y=0.50*(1)------> y=0.50-----> is correct

for x=2

y=0.50*(2)------> y=1.00-----> is correct

for x=3

y=0.50*(3)------> y=1.50-----> is correct

for x=5

y=0.50*(5)------> y=2.50-----> is correct

4 0

3 years ago

Read 2 more answers

Karen and Bill collect coins. Karen has x coins. Bill has 9 coins fewer than two times the number of coins Karen has. Write and

Eduardwww [97]

Karen = x
bill = 2x - 9
total = x + 2x - 9 = 3x - 9

6 0

2 years ago

Find the flow rate through a tube of radius 6 cm, assuming that the velocity of fluid particles at a distance r cm from the cent

vaieri [72.5K]

Answer:

791.68 cm/s

Step-by-step explanation:

The volume flow rate can be interpreted as the integral of fluid velocity over area

$\dot{V} = \int\limits^6_0 {v(r) 2\pi r} \, dr\\\dot{V} = 2\pi\int\limits^6_0 {(25-r^2)r} \, dr\\\dot{V} = 2\pi\int\limits^6_0 {25r-r^3} \, dr\\\\\dot{V} = 2\pi[12.5r^2 - r^4/4]_0^6\\\dot{V} = 2\pi(12.5*6^2 - 6^4/4 - 12.5*0 - 0)\\\dot{V} = 2\pi*126 = 791.68 cm/s$

7 0

3 years ago