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maks197457 [2]
2 years ago
15

Six statements were made about the figures below. Record the letter of the 3D figure that matches each student’s statement.

Mathematics
1 answer:
nekit [7.7K]2 years ago
8 0

The answer will be 1-sphere,2-Triangular pyramid,3-cube,4-Triangular prism,5-Cylinder,6-Rectangular pyramid

<h3>What are the figures?</h3>

The figures can be sphere, cylinder, cube,pyramid,prism etc.We have to match the description with their correct solid name.

figure where all points are the same distance from the center, is called sphere.

1-sphere

figure made of all triangles.That figure is called a triangular pyramid.

2-Triangular pyramid

figure with six congruent squares is called a cube.

3-cube

figure with two triangle bases and rectangles for the rest of its faces.That figure is called a triangular prism.

4-Triangular prism

figure with circular bases is called a cylinder.

5-Cylinder

figure with a rectangle as a base and a triangle for the rest of its bases.

That figure is called a rectangular pyramid.

6-Rectangular pyramid

To know more about the Figures follow

brainly.com/question/15981553

#SPJ1

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If f(x) = -(x - 1) then does f(-2)=3?​
Aneli [31]

Answer:

Yes thats correct

Step-by-step explanation:

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3 years ago
Y=(3/5)x -3 and 5y =3x-10 these lines are and the second question?
azamat
Look at the first line:  y = (3/5)x - 3.  What happens if you multiply each term by 5, to eliminate the fraction?

5y = 3x - 3

Compare this to the second equation, 

5y - 3x = -10, or 5y = 3x - 10.

The coefficients of x and y (as 3 and 5 here) determine the slope of a straight line.  Since 5y = 3x is present in both equations, the two lines MUST be parallel.



y = 4
4y = 6   =>   y = 6/4

y+4 and y =3/2 are both horizontal lines.  Since they are horiz., they are parallel.


4 0
3 years ago
The product of negative six and the sum of q plus two is twenty.
marshall27 [118]

Answer:

q=24

Step-by-step explanation:

2+-6=-4

20+4=24

The 24 is the sum of the 20 and to turn the negative into 0 plus the 20 equals 24

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3 years ago
A magazine has 16 pages; 5 pages are short stories, and 6 pages are
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Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1
Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

x'=\frac{dx}{dt}

a)  x’=t–sin(t),  x(0)=1

dx=(t-sint)dt

Apply integral both sides:

\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k

where k is a constant due to integration. With x(0)=1, substitute:

1=0+cos0+k\\\\1=1+k\\k=0

Finally:

x=\frac{t^2}{2} +cos(t)

b) x’+2x=4; x(0)=5

dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\

Completing the integral:

-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}

Solving the operator:

-\frac{1}{2}ln(4-2x)=t+k

Using algebra, it becomes explicit:

x=2+ke^{-2t}

With x(0)=5, substitute:

5=2+ke^{-2(0)}=2+k(1)\\\\k=3

Finally:

x=2+3e^{-2t}

c) x’’+4x=0; x(0)=0; x’(0)=1

Let x=e^{mt} be the solution for the equation, then:

x'=me^{mt}\\x''=m^{2}e^{mt}

Substituting these equations in <em>c)</em>

m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)

Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}

Finally:

x=\frac{1}{2} sin(2t)

7 0
3 years ago
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