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3 years ago

WILL GIVE BRAINLIEST!!!

Mathematics

1 answer:

Mnenie [13.5K]3 years ago

8 0

The answer to your question is 34.13%

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Find the equation in standard form. of the line that passes through (-2,4) and (1,-2)

blagie [28]

-3 will be your slope and your y-intercept will be -2... equation i: y=-3x-2

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3 years ago

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A 2 kg metal cylinder is supplied with 1600J of energy to heat it from 5°C to 13°C. What is the specific heat capacity of the me

Neporo4naja [7]

Answer:

$c = 100\,\frac{J}{kg\cdot ^{\circ}C}$

Step-by-step explanation:

According to the First Law of Thermodynamics, the heat received by the metal cylinder is equal to the change in the internal energy. That is:

$Q = \Delta U$

$Q = m \cdot c \cdot \Delta T$

The specific heat is clear in the previous expression and finally computed:

$c = \frac{Q}{m\cdot \Delta T}$

$c = \frac{1600\,J}{(2\,kg)\cdot (13^{\circ}C-5^{\circ}C)}$

$c = 100\,\frac{J}{kg\cdot ^{\circ}C}$

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3 years ago

Can I have these questions answered,

weeeeeb [17]

1. 37.0952380952 or if rounded 38
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3 years ago

The environmental science club is printing T-shirts for it's 15 members. The printing company charges a certain amount for each

alexandr402 [8]

Answer:

Step-by-step explanation:

Let $X$ be the cost of each shirt. From the information provided in the problem, we can setup the following equation:

$15X + 20 = 162.50$

Solving for $X$ will give us the answer:

$15X = 142.50$

$X = 9.5$

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3 years ago

Analyzing a Graph In Exercise, analyze and sketch the graph of the function. Lable any relative extrema, points of inflection, a

Sladkaya [172]

Answer:

$y=(\ln{x})^2$

point of extremity: (1,0)

vertical asymptote: along the y-axis (x = 0)

point of inflection: (e,1)

Solution:

Although all these points can be directly observed from the graph below, but these are the analytical solutions if you're curious!

1) Extreme point can be found by differentiating 'y' once and equating to zero. solving for x:

$\dfrac{dy}{dx}=\dfrac{dy}{dx}((\ln{x})^2)$

$\dfrac{dy}{dx}=2\ln{x}\left(\dfrac{1}{x}\right)$

substitute dy/dx = 0, and solve for x

$0=2\ln{x}\left(\dfrac{1}{x}\right)$

$0=2\ln{x}$

$x=e^0$

$x=1$

use this value of x back in y, to find the y-coordinate of the extreme point

$y=(\ln{1})^2$

$y=0$

The extreme point = (1,0)

2) Differentiate y twice to find the inflection point.

$\dfrac{dy}{dx}=2\ln{x}\left(\dfrac{1}{x}\right)$

$\dfrac{d^2y}{dx^2}=2\ln{x}\left(-\dfrac{1}{x^2}\right)+\left(\dfrac{1}{x}\right)\left(\dfrac{2}{x}\right)$

$\dfrac{d^2y}{dx^2}=\dfrac{2}{x^2}\left(-\ln{x}+1}\right)$

substitute d2y/dx2 = 0, and solve for x

$0=\dfrac{2}{x^2}\left(-\ln{x}+1}\right)$

$0=-\ln{x}+1$

$\ln{x}=1$

$x = e$

use this value of x back in y, to find the y-coordinate of the inflection point

$y=(\ln{e})^2$

$y=1$

The extreme point = (e,1)

6 0

3 years ago