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2 years ago

Select the correct answer. Which relationship does the graph represent?

Mathematics

1 answer:

vladimir2022 [97]2 years ago

8 0

Answer:

Step-by-step explanation:

perimeter of square = 4 length of its sides

x represents the side lengths

y represents the perimeter of the a square

y = ax + b

a = slope = y2-y1 / x2 -x1

= 8-4 / 2-1 = 4

y = 4x + b

from any point (1,4 )

4 = 4 .(1) + b

b= 0

y = 4x

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3 years ago

If a(5,3),b(0,0) and c(-3,5) are the vertices of a triangle then the value of AB.BC is

statuscvo [17]

The dot product associated to sides AB and BC is 0.

<h3>What is the dot product generated by two sides of a triangle?</h3>

In this question we must find the dot product of two vectors associated with two surrounding sides of a triangle. First, we need to find the vectors associated to sides AB and BC:

Side BA

BA = (5, 3) - (0, 0)

BA = (5, 3)

Side BC

BC = (- 3, 5) - (0, 0)

BC = (- 3, 5)

By definition of dot product, the result associated to the two sides of the triangle is:

BA • BC = (5, 3) • (- 3, 5)

BA • BC = (5) · (- 3) + (3) · (5)

BA • BC = - 15 + 15

BA • BC = 0

The dot product associated to sides AB and BC is 0.

<h3>Remark</h3>

The statement is poorly formatted, the correct form is shown below:

If A(x, y) = (5, 3), B(x, y) = (0, 0) and C(x, y) = (- 3, 5) are the vertices of a triangle then the value of the dot product between vectors AB and BC is:

To learn more on dot product: brainly.com/question/14455586

#SPJ1

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2 years ago

A tank contains 1080 L of pure water. Solution that contains 0.07 kg of sugar per liter enters the tank at the rate 7 L/min, and

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(a) Let $A(t)$ denote the amount of sugar in the tank at time $t$ . The tank starts with only pure water, so $\boxed{A(0)=0}$ .

(b) Sugar flows in at a rate of

(0.07 kg/L) * (7 L/min) = 0.49 kg/min = 49/100 kg/min

and flows out at a rate of

(A(t)/1080 kg/L) * (7 L/min) = 7A(t)/1080 kg/min

so that the net rate of change of $A(t)$ is governed by the ODE,

$\dfrac{\mathrm dA(t)}[\mathrm dt}=\dfrac{49}{100}-\dfrac{7A(t)}{1080}$

$A'(t)+\dfrac7{1080}A(t)=\dfrac{49}{100}$

Multiply both sides by the integrating factor $e^{7t/1080}$ to condense the left side into the derivative of a product:

$e^{\frac{7t}{1080}}A'(t)+\dfrac7{1080}e^{\frac{7t}{1080}}A(t)=\dfrac{49}{100}e^{\frac{7t}{1080}}$

$\left(e^{\frac{7t}{1080}}A(t)\right)'=\dfrac{49}{100}e^{\frac{7t}{1080}}$

Integrate both sides:

$e^{\frac{7t}{1080}}A(t)=\displaystyle\frac{49}{100}\int e^{\frac{7t}{1080}}\,\mathrm dt$

$e^{\frac{7t}{1080}}A(t)=\dfrac{378}5e^{\frac{7t}{1080}}+C$

Solve for $A(t)$ :

$A(t)=\dfrac{378}5+Ce^{-\frac{7t}{1080}}$

Given that $A(0)=0$ , we find

$0=\dfrac{378}5+C\implies C=-\dfrac{378}5$

so that the amount of sugar at any time $t$ is

$\boxed{A(t)=\dfrac{378}5\left(1-e^{-\frac{7t}{1080}}\right)}$

(c) As $t\to\infty$ , the exponential term converges to 0 and we're left with

$\displaystyle\lim_{t\to\infty}A(t)=\frac{378}5$

or 75.6 kg of sugar.

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