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stepladder [879]
3 years ago
13

Which of the following solids can be sliced horizontally or vertically to create a triangular cross section?

Mathematics
2 answers:
Blizzard [7]3 years ago
7 0

Answer:

What the other guy said

ehidna [41]3 years ago
4 0

Answer:

Cone, pyramid and triangular cube

Step-by-step explanation:

Ⓗⓘ ⓣⓗⓔⓡⓔ

The cone, pyramid and triangular cube thing

(っ◔◡◔)っ ♥ Hope this helped! Have a great day! :) ♥

Please, please give brainliest, it would be greatly appreciated, I only a few more before I advance, thanks!

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maggie's brother is 3 years younger than twice her age. the sum of their ages is 24. write solve an equation to determine maggie
prohojiy [21]
Maggie =x
brother = 2x-3
x+2x-3= 24
27 /3= 3 x/3
x= 9 years
7 0
3 years ago
Max spent $53 and now has no money left. He had $ before his purchase.
Yuri [45]
Can you restate it or rewrite it? It doesn't have enough for me to answer this.
8 0
2 years ago
Y=3x-5 y=6x-8 show all steps and write the solution
m_a_m_a [10]

Step 1: Set the two equations equal to each other and solve for x.

3x -5 = 6x - 8

3x + (-5+5) = 6x -8 + 5

(3x - 6x) = (6x - 6x) - 3

-3x/-3  = -3/-3

x = 1

Step 2: To solve for y take one of the given equation of your choice (for the purpose of this explanation I will only do y = 3x - 5) and replace x with 1, then solve for y

y = 3(1) - 5

y = 3 - 5

y = -2

(1,-2)

Check:

-2 = 3(1) - 5 ---> - 2 = -2

-2 = 6(1) - 8 ---> -2 = -2

Hope this helped!

5 0
2 years ago
Read 2 more answers
Minimum and maximum value of 2n^2+5n-25=0
Alborosie
Equations don't have minimum or maximum, functions do.

Function y=2n^2+5n-25 has minimum -28.125, has no maximum.
8 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
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