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3 years ago

Can someone please help me

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

3 0

Answer:

Howdy! Let me see if I can be of assistance.

This is a triangular prism that we must solve for, and there are a few steps that can guarantee you land on the right answer.

1) Plug the base and height numbers into the formula, where V is volume:

V= 1/2 × base × height

which becomes

V= 1/2 × 6 × 10

Note that either of the base numbers will work, since the ratio is the same.

2) multiply both of those numbers by 1/2 and you will have your answer! I leave that to your capable hands.

P.S. remember to put the answer down as centimeters squared

Happy learning, friend!

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Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of

Simora [160]

Answer:

The region between a chord and either of its arcs is called a segment the circle.

Angles in the same segment of a circle are equal.

Step-by-step explanation:

Let bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F .

Now from figure,

∠D = ∠EDF

∠D = ∠EDA + ∠ADF

Since ∠EDA and ∠EBA are the angles in the same segment of the circle.

So ∠EDA = ∠EBA

Hence ∠D = ∠EBA + ∠FCA

Again ∠ADF and ∠FCA are the angles in the same segment of the circle.

hence ∠ADF = ∠FCA

Again since BE is the internal bisector of ∠B and CF is the internal bisector of ∠C

So ∠D = ∠B/2 + ∠C/2

Similarly

∠E = ∠C/2 + ∠A/2

and

∠F = ∠A/2 + ∠B/2

Now ∠D = ∠B/2 + ∠C/2

=∠D = (180 - ∠A)/2

(∠A + ∠B + ∠C = 180)

∠D = 90 - ∠A/2

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8 0

3 years ago

Please help me!!!! with both

loris [4]

Answer:

Part 11) $a_n=a_{n-1}-16$ , $a_1=17$

Part 2) 4+3i

Step-by-step explanation:

Part 11) write a recursive rule for the sequence

$17,1,-15,-31,...$

Let

$a_1=17\\a_2=1\\a_3=-15\\a_4=-31$

we know that

$a_2-a_1=1-17=-16$ ----> $a_2=a_1-16$

$a_3-a_2=-15-1=-16$ ----> $a_3=a_2-16$

$a_4-a_3=-31-15=-16$ ----> $a_4=a_3-16$

This is an arithmetic sequence

In an Arithmetic Sequence the difference between one term and the next is a constant, and this constant is called the common difference (d).

In this problem the common difference is equal to $d=-16$

therefore

A recursive rule for the sequence is

$a_n=a_{n-1}+d$

substitute the value of d

$a_n=a_{n-1}-16$

where

$a_1=17$

Part 12) What is the complex conjugate of 4-3i?

we know that

The complex conjugate of a complex number is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign

we have that

the real part of the given number is 4 and the the imaginary part is -3i

the imaginary part equal in magnitude but opposite in sign is +3i

therefore

the complex conjugate of the given number is equal to

$4+3i$

6 0

3 years ago