What is the correct answer someone pls help!?????

Which of the following is not part of the Triangle Proportionality Theorem?

natka813 [3]

Answer:

A. AA Similarity Theorem . True

B. Two sides divided proportionally. True

C. A parallel line to one side of a triangle. True

D. A perpendicular line to one side of a triangle. False

Step-by-step explanation:

The Triangle Proportionality Theorem says that if a line is parallel to one side of a triangle, then it splits the other two sides into proportional sections.

Now, here the given statements are:

A. AA Similarity Theorem

TRUE, as the angles opposite to equal sides are equal.

And if lines are divided proportionally, both the triangles are Similar by AA similarity theorem.

B. Two sides divided proportionally.

TRUE,as the other two sides of the triangle are divided proportionally by the parallel line.

C. A parallel line to one side of a triangle.

TRUE, as the parallel line divides the other two non- parallel sides in equal proportions.

D. A perpendicular line to one side of a triangle.

FALSE, this part is not required anywhere in the given theorem result.

3 0

3 years ago

(c). It is well known that the rate of flow can be found by measuring the volume of blood that flows past a point in a given tim

aleksklad [387]

(i) Given that

$V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr$

when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then

$V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}$

and this is a volume so it must be reported with units of cm³.

In Mathematica, you can first define the velocity function with

v[r_] := 0.30 - 3.33r^2

and additionally define the volume function with

V[R_] := Integrate[2 Pi v[r] r, {r, 0, R}]

Then get the desired volume by running V[0.30].

(ii) In full, the volume function is

$\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

Compute the integral:

$V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

$V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr$

$V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R$

$V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)$

$V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}$

In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

5 0

3 years ago

Solve by using trial and error method. 3x = 16

Paladinen [302]

Answer:

5.3...

..............

4 0

2 years ago

What is -15x=-240?????????

Vlad [161]

You divide -15 on both sides to keep the x alone
-240/ -15=16
A negative divided by a negative is a positive so x=16

4 0

3 years ago

Read 2 more answers

A fisherman separated of his catch into 2 equal bins.

ohaa [14]

This question doesn't make any sense please reword and i will do my best to solve.

5 0

3 years ago