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3 years ago

In triangle ABC shown below, side AB is 8 and side AC is 6:

Mathematics

2 answers:

ira [324]3 years ago

3 0

The answer is C because AD is half of AB and AE is half of AC.

Burka [1]3 years ago

3 0

Answer: The correct option is

(D) Segment AD is 4, and segment AE is 3.

Step-by-step explanation: In the given figure, we are shown a triangle ABC with side AB measures 8 units and AC measures 6 units.

We are to select the statement that is needed to prove that segment DE is half the length of segment BC.

Midpoint Theorem :

The line joining the mid points of any two sides of a triangle is parallel to the third side and is half of the third side.

In triangle ABC, applying the midpoint theorem,

for DE to be half of BC, D must be the midpoint of AB and E must be the midpoint of AC.

If so, then

$AD=\dfrac{1}{2}\times AB=\dfrac{1}{2}\times8=4,\\\\\\AE=\dfrac{1}{2}\times AC=\dfrac{1}{2}\times 6=3.$

Thus, for DE to be half of BC, segment AD is 4 and segment AE is 3.

Option (D) is CORRECT.

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In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago

Prove that A is in the subset B if P(A) is in the subset P(B)

natima [27]

Show that if p(a) ⊂ p(b) then a ⊂ b.

I will assume p() means power set. 

proof: let x∈a, then {x} ∈ p(a) and so by hypothesis {x} ∈ p(b). However {x} could not be in p(b) unless x∈b. This shows that each element of a is an element of b and hence a ⊂ b.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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3 years ago

What I Can Do

tankabanditka [31]

Answer:

5Stetpplanation:

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3 years ago

Can a. True b. False results be in a probability distribution

Alexxandr [17]

I think the answers True hope this helps

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3 years ago

A car is traveling at a speed of

riadik2000 [5.3K]

108 kilometers is 108,000 meters

108,000 ÷ 60 = 1800 meters per minute

1800 ÷ 60 = 30

5 0

2 years ago