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2 years ago

Could you help me understand cross sections of three-dimenstional object its harder than it sound.

1 answer:

5 0

A section, or cross-section, is a view of a 3-dimensional object from the position of a plane through the object. A section is a common method of depicting the internal arrangement of a 3-dimensional object in two dimensions. It is often used in technical drawing and is traditionally crosshatched.

Cross sections of three-dimensional objects are two-dimensional shapes of various sizes. They may be parallel to a side or base of the object or at an angle to these surfaces. A cross section may resemble the shape of the object’s side or base, or it may have a completely different shape.

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what is the perimeter of the triangle. i do not know how to start this problem. any help will be greatly appreciated :))

xz_007 [3.2K]

Given:

Point F,G,H are midpoints of the sides of the triangle CDE.

$FG=9,GH=7,CD=24$

To find:

The perimeter of the triangle CDE.

Solution:

According to the triangle mid-segment theorem, the length of the mid-segment of a triangle is always half of the base of the triangle.

FG is mid-segment and DE is base. So, by using triangle mid-segment theorem, we get

$\dfrac{1}{2}DE=FG$

$DE=2(FG)$

$DE=2(9)$

$DE=18$

GH is mid-segment and CE is base. So, by using triangle mid-segment theorem, we get

$\dfrac{1}{2}CE=GH$

$CE=2(GH)$

$CE=2(7)$

$CE=14$

Now, the perimeter of the triangle CDE is:

$Perimeter=CD+DE+CE$

$Perimeter=24+18+14$

$Perimeter=56$

Therefore, the perimeter of the triangle CDE is 56 units.

7 0

2 years ago

Curtis bought a skateboard which he later sold at a 30% loss for £14. How much did Curtis pay for

Norma-Jean [14]

Answer:

18 pounds if I'm correct

3 0

2 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

vodka [1.7K]

Answer:

a) $v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

b) $0$

c) $a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

$x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10$

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

$v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

$v(t) = 3t^2 -12t +9$

We can factorize this function as $v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)$

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is $0\leq t\leq 10$ we need to analyze the following intervals:

$0< t$

For this case if we select two values let's say 0.25 and 0.5 we see that

$v(0.25) =6.1875, v(0.5)=3.75$

And we see that for $a=0.5 >0.25=b$ we have that f(b)>f(a) so then the function is decreasing on this case.

$1$

We have a minimum at t=2 since at this value w ehave the vertex of the parabola :

$v_x =-\frac{b}{2a}= -\frac{-12}{2*3}= -2$

And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2

$2$

We see that the function would be increasing.

$3$

For this interval we will see that for any two points a,b with $a>b$ we have $f(a)>f(b)$ for example let's say a=3 and b =4

$f(a=3) =0 , f(b=4) =9 , f(b)>f(a)$

The particle is moving to the right then the velocity is positive so then the answer for this case is: $0$

Part c

$a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

5 0

3 years ago

Need answer asap.thanks

Marrrta [24]

Answer:

$0.46=\frac{23}{50}$

Step-by-step explanation:

To write any decimal as a fraction you divide by 1 and multiply by a number (ranging from 10, 100, 1000 etc.) that will make 0.46 a whole number, this will explain:

Let x = $\frac{0.46}{1}$

10x = $10*\frac{0.46}{1} =\frac{4.6}{10}$

100x = $100*\frac{0.46}{1}=\frac{46}{100}$ this is our perfect fraction, now we simplify later

100x - 10x = $\frac{46}{100} -\frac{4.6}{10}$

90x = $\frac{46}{100} -\frac{4.6}{10} =0$ this is to confirm both fractions are equal

x is the same as $\frac{0.46}{1}$ as $\frac{4.6}{10}$ as $\frac{46}{100}$ but here x = $\frac{46}{100}$ because a fraction has to have no decimals.

So 0.46 is equal any of these values, as a fraction, on the other hand, it's improperly equal to $\frac{46}{100} = \frac{23}{50}$ here I divided by 2 to bring down the proper fraction. (fraction at its simplest form)

6 0

3 years ago

A brand new filled jar of salsa is 11 cm tall and a has a radius of 6 cm. Kelly eats some of the salsa and the salsa in the jar

Sonbull [250]

Before we begin please note that a jar is always cylindrical in shape if not mentioned otherwise. Thus, we will take a cylindrical jar in this case.

Since Kelly eats some of the salsa and the salsa in the jar is now 6 cm high, we know that Kelly ate the amount of Salsa whose volume is missing from the jar.

To find the missing volume on the Salsa jar all that we need to do is to find the height of the missing volume which is given to us as h= 6 cm. We already know that the radius of the jar, r=6 cm. Thus, the amount of Salsa Kelly has eaten is:

$V=\pi(6)^2\times 6\approx678.6\approx679$ $cm^3$

Thus, the third option is the correct answer

8 0

2 years ago

Read 2 more answers

Could you help me understand cross sections of three-dimenstional object its harder than it sound.​

Could you help me understand cross sections of three-dimenstional object its harder than it sound.