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2 years ago

What is the value of x? Enter your answer in the box. x =

Mathematics

1 answer:

qaws [65]2 years ago

7 0

Answer:

Step-by-step explanation: its yo mama :)

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Find the area of each figure. Round to the nearest tenth if necessary.

natita [175]

Answer:

Solution given:

diameter(d)=6mm

base(b)=8mm

height (h)=5mm

Area of figure=area of parallelogram +area of semi circle

base*height+½π(d/2)²
8*5+½*π×(6/2)²
40+14.14
54.4mm²
Area :54.14mm²

for triangle

base[b]=6ft

height(h)=9ft

for square

length[l]=9ft

Area of figure=area of square +area of triangle

=l²+½*b*h
=9²+½*6*9
=81+27
=108ft²
Area: 108ft²

6 0

2 years ago

7 What is the value of the expression -3x^2 y +4x when x = -4 and y = 2

love history [14]

-3x^2 y +4x
 =-3(-4)^2 (2) + 4(-4)
=-96-16
=-112

7 0

3 years ago

What is the input in a scatter plot

Charra [1.4K]

Answer:

the X variable

Step-by-step explanation:

it's the same as any other function

3 0

2 years ago

The formula for the area,A, of a trapezoid is shown below, where ݄H represents its height and b1 and b2

emmainna [20.7K]

One equivalent is $A=h( \frac{ b_{1}+b_{2} }{2})$

Another example is $A= \frac{(b_{1}+b_{2})(h)}{2}$

4 0

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