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3 years ago

Please help me on #2! Could you also check the other numbers! Thank you so much!

Mathematics

1 answer:

alexira [117]3 years ago

8 0

Dani is correct. The equation can be written as y = 3x+50, which is a linear equation in the form y = mx+b. We have m = 3 as the slope and b = 50 as the y intercept

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x is the number of square feet, y is the total cost in dollars

3*x or 3x repesents multiplying the cost per square foot ($3) by the number of square feet (x). For example, if you have 10 sq ft, then so far the cost would be 3*x = 3*10 = 30 dollars.

This result is then added to 50 to get the total cost. We say that 3x is the variable cost, as it changes or varies depending on how large x gets. Additionally, the 50 is the fixed cost because this cost stays the same no matter what x is

total cost = variable cost + fixed cost

y = 3x + 50

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Plz help me with this question

Montano1993 [528]

Answer:

Step-by-step explanation:

Remark

There are probably a couple of ways that you could do this. I'm going to pick the most obvious way.

Givens.

KN = 2.5

NO = 10

LM = 3

MP = x

Equation

NO/KN = PM / LM

Solution

10 / 2.5 = x / 3 Cross multiply

2.5x = 10 * 3 Combine

2.5x = 30 Divide by 2.5

x = 30/2.5

x = 12

That looks like it should be the answer, and it almost is, but you are asked for LP.

LP = LM + MP

LP = 3 + 12

LP = 15

The answer is 15

7 0

3 years ago

Carlos has a box of coins that he uses when playing poker

Fofino [41]

Answer:

so far i got 8 quarters 14 dimes and 13 pennies

Step-by-step explanation:

ill update you if i got any closer

6 0

3 years ago

Three vertices of parallelogram JKLM are J(1, 4), K(5, 3), and L(6,−3). Find the coordinates of vertex M.

Andrej [43]

Answer:

coordinates of vertex M is (x, y) = (2, -2)

Step-by-step explanation:

Since JKLM is a parallelogram, this implies that JK parallel to LM and KL parallel to JM. This means that

Slope of JK = slope of LM

$\frac{3-4}{5-1} =\frac{-3-y}{6-x} \\y=-\frac{1}{4}x-\frac{3}{2} ....(i)$

And

Slope of KL = slope of JM

$\frac{3-\left(-3\right)}{5-6}=\frac{4-y}{1-x}\\y=10-6x...(ii)$

From equation (i) and (ii) we get

$-\frac{1}{4}x-\frac{3}{2} =10-6x$

$-\frac{23x}{4}=-\frac{23}{2}$

$-23x=-46$

$x=2$

Put the value of x in equation (ii) we get

$y=10-6(2)\\y=-2$

So, the coordinates of vertex M is (x, y) = (2, -2).

5 0

3 years ago

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2

Alika [10]

As the ladder is pulled away from the wall, the area and the height with the

wall are decreasing while the angle formed with the wall increases.

The correct response are;

(a) The velocity of the top of the ladder = 1.5 m/s downwards

(b) The rate the area formed by the ladder is changing is approximately -75.29 ft.²/sec

Reasons:

The given parameter are;

Length of the ladder, l = 25 feet

Rate at which the base of the ladder is pulled, $\displaystyle \frac{dx}{dt}$ = 2 feet per second

(a) Let y represent the height of the ladder on the wall, by chain rule of differentiation, we have;

$\displaystyle \frac{dy}{dt} = \mathbf{\frac{dy}{dx} \times \frac{dx}{dt}}$

25² = x² + y²

y = √(25² - x²)

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} \sqrt{25^2 - x^2} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}$

Which gives;

$\displaystyle \frac{dy}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times \frac{dx}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2$

$\displaystyle \frac{dy}{dt} = \mathbf{ \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2}$

When x = 15, we get;

$\displaystyle \frac{dy}{dt} = \frac{15 \times \sqrt{625-15^2} }{15^2- 625}\times2 = \mathbf{-1.5}$

The velocity of the top of the ladder = 1.5 m/s downwards

When x = 20, we get;

$\displaystyle \frac{dy}{dt} = \frac{20 \times \sqrt{625-20^2} }{20^2- 625}\times2 = -\frac{8}{3} = -2.\overline 6$

The velocity of the top of the ladder = $\underline{-2.\overline{6} \ m/s \ downwards}$

When x = 24, we get;

$\displaystyle \frac{dy}{dt} = \frac{24 \times \sqrt{625-24^2} }{24^2- 625}\times2 = \mathbf{-\frac{48}{7}} \approx -6.86$

The velocity of the top of the ladder ≈ -6.86 m/s downwards

(b) $\displaystyle The \ area\ of \ the \ triangle, \ A =\mathbf{\frac{1}{2} \cdot x \cdot y}$

Therefore;

$\displaystyle The \ area\ A =\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}$

$\displaystyle \frac{dA}{dx} = \frac{d}{dx} \left (\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}\right) = \mathbf{\frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250}}$

$\displaystyle \frac{dA}{dt} = \mathbf{ \frac{dA}{dx} \times \frac{dx}{dt}}$

Therefore;

$\displaystyle \frac{dA}{dt} = \frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250} \times 2$

When the ladder is 24 feet from the wall, we have;

x = 24

$\displaystyle \frac{dA}{dt} = \frac{(2 \times 24^2- 625)\cdot \sqrt{625-24^2} }{2\times 24^2 - 1250} \times 2 \approx \mathbf{ -75.29}$

The rate the area formed by the ladder is changing, $\displaystyle \frac{dA}{dt}$ ≈ -75.29 ft.²/sec

$\displaystyle sin(\theta) = \frac{x}{25}$

$\displaystyle \theta = \mathbf{arcsin \left(\frac{x}{25} \right)}$

$\displaystyle \frac{d \theta}{dt} = \frac{d \theta}{dx} \times \frac{dx}{dt}$

$\displaystyle\frac{d \theta}{dx} = \frac{d}{dx} \left(arcsin \left(\frac{x}{25} \right) \right) = \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625}}$

Which gives;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-x^2} }{x^2 - 625}\times \frac{dx}{dt}= \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625} \times 2}$

When x = 24 feet, we have;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-24^2} }{24^2 - 625} \times 2 \approx \mathbf{ 0.286}$

Rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall is $\displaystyle \frac{d \theta}{dt}$ ≈ 0.286 rad/sec

Learn more about the chain rule of differentiation here:

brainly.com/question/20433457

3 0

3 years ago

In the function, 9 (2-) = 9 - 5x, what is the value of g (3)?

vekshin1

Answer: 53

Step-by-step explanation:

3 0

3 years ago