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2 years ago

O

Mathematics

1 answer:

erastovalidia [21]2 years ago

6 0

Answer:

D .

Sorry if i'm wrong.

$x(2 {x}^{2} + 4x - 1 )\\ = 2 {x}^{3} + 4 {x}^{2} - x$

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What is the area of a sector with a central angle of (2pi/3) radians and a diameter of 12 in?

Stels [109]

Answer:

The area of the sector is $37.68\ in^{2}$

Step-by-step explanation:

step 1

Find the area of the circle

The area of the circle is equal to

$A=\pi r^{2}$

we have

$r=12/2=6\ in$ ----> the radius is half the diameter

substitute

$A=(3.14)(6)^{2}$

$A=113.04\ in^{2}$

step 2

Find the area of a sector with a central angle of (2pi/3)

Remember that

The area of $113.04\ in^{2}$ subtends a central angle of $2\pi \ radians$

by proportion

Let

x----> the area of the sector

$\frac{2\pi}{113.04}=\frac{(2\pi/3)}{x}\\ \\x=113.04*(2\pi/3)/(2\pi)\\ \\x=37.68\ in^{2}$

8 0

3 years ago

Tamara wants to calculate the height of a tree outside her home. Using her clinometer, she measures the angle from 35 feet away

serg [7]

Answer:

71.32 feet

Step-by-step explanation:

We set up a trigonometric function to get height of tree

Distance ,x = 35 feet away from tree base

Angle = 62⁰ at eye level

Y = tan(angle) * x

= Y = tan(62⁰) x 35

Y = 1.881 x 35

Y = 65.82 feet

We add Tamara's height to the total height of the tree

= 65.82feet + 5.5 feet

= 71.32 feet is the height of the tree

3 0

3 years ago

Match solutions and differential equations. Note: Each equation may have more than one solution. Select all that apply. (a) 7y''

Nonamiya [84]

Answer:

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

Step-by-step explanation:

a.7 y''-7 y =0

Auxillary equation

$D^2-1=0$

$(D-1)(D+1)=0$

D=1,-1

Then , the solution of given differential equation

$y=e^x,y=e^{-x}$

2. $7x^2y''+14xy'-14 y=0$

Y= $y=x^3$

$y=3x^2$

$y''=6x$

Substitute in the given differential equation

$42x^3+42x^3-14x^3\neq 0$

Hence, $x^3$ is not a solution of given differential equation

$e^x,e^{-x}$ are also not a solution of given differential equation.

y= $x^{-2}$

$y'=-2x^{-3}$

$y''=6x^{-4}$

Substitute the values in the differential equation

$7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}$

= $42x^{-2}-28x^{-2}-14x^{-2}=0$

Hence, $x^{-2}$ is a solution of given differential equation.

c. $7x^2y''-42y=0$

$y=x^3$

$y'=3x^2$

$y''=6x$

Substitute the values in the differential equation

$42x^3-42x^3=0$

Hence, $x^3$ is a solution of given differential equation.

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

6 0

3 years ago

Read 2 more answers

Luba_88 [7]

Answer:

Step-by-step explanation:

$y\leq\frac{1}{2}x+2 \\\\1\leq \frac{1}{2}(0)+2\\\\1\leq 0+2\\\\1\leq 2$

This statement is true

Hope this helps!

3 0

3 years ago

I need help please!!!

MAVERICK [17]

Answer:

1). 10. 2). 9. 3). 2(square root)14. 4). (square root)6. 5). right triangle. 6). not a right triangle. 7). 13. 8). 3(square root)34. 9). acute. 10). acute. 11). right. 12). obtuse

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers