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3 years ago

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How many triangles can be constructed with sides measuring 5 m, 16 m, and 5 m?

1 answer:

lidiya [134]3 years ago

8 0

Q. How many triangles can be constructed with sides measuring 5 m, 16 m, and 5 m?

Solution:

Here we are given with the sides of the triangle as 5m, 16m and 5.

As the Triangle inequality we know that

The sum of the length of the two sides should be greater than the length of the third side. But this inequality fails here.

Hence no triangle can be made.

So the correct option is None.

Q.How many triangles can be constructed with sides measuring 6 cm, 2 cm, and 7 cm?

Solution:

Here we are given with the sides of the triangle as 6m, 2m and 7m.

As the Triangle inequality we know that

The sum of the length of the two sides should be greater than the length of the third side. The given values follows the triangle inequality.

Hence one triangle can be formed.

So the correct option is one.

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1.a) Convert 81° into grade.<br>10

Kazeer [188]

Step-by-step explanation:

I hope it helps man this much .

6 0

3 years ago

What is the converse of the following statement? If a number is odd, it is not divisible by two.

tensa zangetsu [6.8K]

Converse in discrete mathematics is switching the hypothesis and conclusion of a conditional statement.

So the correct answer will be that If a number is not divisible by two, then it is an odd number. Which is option D.

The math converse of a statement switches the if and then, resulting in a statement that may or may not be true so we have no guarantee that the preposition is true or not.

8 0

3 years ago

This is a "water tank" calculus problem that I've been working on and I would really appreciate it if someone could look at my w

Sedaia [141]

Part A

Everything looks good but line 4. You need to put all of the "2h" in parenthesis so the teacher will know you are squaring all of 2h. As you have it right now, you are saying "only square the h, not the 2". Be careful as silly mistakes like this will often cost you points.

============================================================

Part B

It looks like you have the right answer. Though you'll need to use parenthesis to ensure that all of "75t/(2pi)" is under the cube root. I'm assuming you made a typo or forgot to put the parenthesis.

dh/dt = (25)/(2pi*h^2)
2pi*h^2*dh = 25*dt
int[ 2pi*h^2*dh ] = int[ 25*dt ] ... applying integral to both sides
(2/3)pi*h^3 = 25t + C
2pi*h^3 = 3(25t + C)
h^3 = (3(25t + C))/(2pi)
h^3 = (75t + 3C)/(2pi)
h^3 = (75t + C)/(2pi)
h = [ (75t + C)/(2pi) ]^(1/3)

Plug in the initial conditions. If the volume is V = 0 then the height is h = 0 at time t = 0
0 = [ (75(0) + C)/(2pi) ]^(1/3)
0 = [ (0 + C)/(2pi) ]^(1/3)
0 = [ (C)/(2pi) ]^(1/3)
0^3 = (C)/(2pi)
0 = C/(2pi)
C/(2pi) = 0
C = 0*2pi
C = 0

Therefore the h(t) function is...
h(t) = [ (75t + C)/(2pi) ]^(1/3)
h(t) = [ (75t + 0)/(2pi) ]^(1/3)
h(t) = [ (75t)/(2pi) ]^(1/3)

Answer:
h(t) = [ (75t)/(2pi) ]^(1/3)

============================================================

Part C

Your answer is correct.
Below is an alternative way to find the same answer

--------------------------------------

Plug in the given height; solve for t
h(t) = [ (75t)/(2pi) ]^(1/3)
8 = [ (75t)/(2pi) ]^(1/3)
8^3 = (75t)/(2pi)
512 = (75t)/(2pi)
(75t)/(2pi) = 512
75t = 512*2pi
75t = 1024pi
t = 1024pi/75
At this time value, the height of the water is 8 feet

Set up the radius r(t) function
r = 2*h
r = 2*h(t)
r = 2*[ (75t)/(2pi) ]^(1/3) .... using the answer from part B

Differentiate that r(t) function with respect to t
r = 2*[ (75t)/(2pi) ]^(1/3)
dr/dt = 2*(1/3)*[ (75t)/(2pi) ]^(1/3-1)*d/dt[(75t)/(2pi)]
dr/dt = (2/3)*[ (75t)/(2pi) ]^(-2/3)*(75/(2pi))
dr/dt = (2/3)*(75/(2pi))*[ (75t)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)

Plug in t = 1024pi/75 found earlier above
dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (75(1024pi/75))/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (1024pi)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*(1/64)
dr/dt = 25/(64pi)
getting the same answer as before

----------------------------

Thinking back as I finish up, your method is definitely shorter and more efficient. So I prefer your method, which is effectively this:
r = 2h, dr/dh = 2
dh/dt = (25)/(2pi*h^2) ... from part A
dr/dt = dr/dh*dh/dt ... chain rule
dr/dt = 2*((25)/(2pi*h^2))
dr/dt = ((25)/(pi*h^2))
dr/dt = ((25)/(pi*8^2)) ... plugging in h = 8
dr/dt = (25)/(64pi)
which is what you stated in your screenshot (though I added on the line dr/dt = dr/dh*dh/dt to show the chain rule in action)

8 0

3 years ago

Would appreciate some help

GalinKa [24]

Answer:

I think its b plz don't get mad I am not so sure

4 0

3 years ago

Read 2 more answers

Given: ΔPSQ, PS = SQ Perimeter of ΔPSQ = 50 SQ – PQ = 1 Find: Area of ΔPSQ

Lana71 [14]

Answer:

120 units²

Step-by-step explanation:

Perimeter = PS + SQ + PQ

50 = SQ + SQ + (SQ -1)

51 = 3SQ

17 = SQ

17 -1 = 16 = PQ

The midpoint of the base is one leg of the right triangle whose other leg is the height of this isosceles triangle. That height is ...

h = √(17² -(16/2)²) = √225 = 15

Then the area is ...

A = (1/2)bh = (1/2)(16)(15) = 120 . . . . . square units

8 0

3 years ago