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2 years ago

_ are __ midsegments of ΔWXY.

Mathematics

2 answers:

iVinArrow [24]2 years ago

3 0

Answer:

The perimeter of Δ WXY is 14.50 cm ⇒ the last answer

Step-by-step explanation:

* Lets explain how to solve the problem

- There is a fact in any triangle; the segment joining the midpoints of

two side of a triangle is parallel to the 3rd side and half its length

* Lets use this fact to solve the problem

- In Δ WXY

∵ Q is the midpoint of WX

∵ R is the midpoint of XY

∵ S is the midpoint of YW

- By using the fact above

∴ QR = 1/2 WY

∴ RS = 1/2 WX

∴ SQ = 1/2 XY

- Lets calculate the length of the sides of Δ WXY

∵ QR = 1/2 WY

∵ QR = 2.93

∴ 2.93 = 1/2 WY ⇒ multiply both sides by 2

∴ WY = 5.86 cm

∵ RS = 1/2 WX

∵ RS = 2.04

∴ 2.04 = 1/2 WX ⇒ multiply both sides by 2

∴ WX = 4.08 cm

∵ SQ = 1/2 XY

∵ SQ = 2.28

∴ 2.28 = 1/2 XY ⇒ multiply both sides by 2

∴ XY = 4.56 cm

- Lets find the perimeter of Δ WXY

∵ The perimeter of Δ WXY = WX + XY + YW

∴ The perimeter of Δ WXY = 5.86 + 4.08 + 4.56 = 14.50

* The perimeter of Δ WXY is 14.50 cm

satela [25.4K]2 years ago

3 0

Answer:

The perimeter of Δ WXY is 14.50 cm

Step-by-step explanation:

edge2020

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The answer is A.NP.

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answer is (-7,-9)

step by step:-

S is the midpoint of RT so ,

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2 years ago

Given that Ray E B bisects ∠CEA, which statements must be true? Select three options. m∠CEA = 90° m∠CEF = m∠CEA + m∠BEF m∠CEB =

FromTheMoon [43]

Answer:

Here is the question attached with.

$m\angle CEA =90 \ (deg)$

$m\angle BEF=135\ (deg)$

$\angle CEF$ is a straight line.

$\angle AEF$ is a right angled triangle.

Options $1,4,5,6$ are correct answers.

Step-by-step explanation:

⇒As $\ ray\ AE$ is ⊥ $FEC$ so it will forms right angled triangle then $m\angle CEA =90\ (deg)$ .

⇒Measure of $\angle BEF =135\ (deg)$ as $\angle BEF =\angle AEB +\angle AEF = (45+90)=135\ (deg)$ as $\angle AEB$ is the bisector of $\angle AEC$ ,meaning that $\angle AEB$ is half of $\angle AEC$ so $\angle AEB = 45\ (deg)$ .

⇒ $\angle CEF$ is a straight line as the angles measure over it is $180\ (deg)$ .

⇒Measure of $\angle AEF = 90\ (deg)$ from linear pair concept.

As $\angle CEA + \angle AEF = 180\ (deg)$ ,plugging the values of $m\angle CEA =90\ (deg)$ we have $\angle AEF = 90\ (deg)$ .

The other two options are false as:

$m\angle CEF=m\angle CEA + m\angle BEF = (90+135)=225$

it is exceeding $180\ (deg)$ whereas $\angle CEF$ is a

straight line.

And $m\angle CEB=2(m\angle CEA)$ is not true.

As $\angle CEA = 90\ (deg)$ and $\angle CEB=45\ (deg)$

So we have total $4$ answers.

The correct options are $1,4,5,6$ .

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The function D(t) defines a traveler's distance from home, in miles, as a function of time, in hours. D(t) = StartLayout enlarge

nalin [4]

- At 2 hours, the traveler is 725 miles from home.

- At 3 hours, the distance is constant, at 880 miles. --> TRUE.

- The total distance from home after 6 hours is 1,062.5 miles --> TRUE.

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The function D(t) is defined as follows:

$D(t) = 300t+125$ for $t< 2.5$

$D(t) = 880$ for $2.5 \leq 3.5$

$D(t) = 75t+612.5$ for $t\leq 6$

Where

t is the time in hours

D(t) is the distance covered, in miles, after t hours

Now let's analyze the different statements:

- The starting distance, at 0 hours, is 300 miles. --> FALSE. In fact, if we substitute t = 0 into the 1st equation, we get

$D(0) = (300)(0)+125 = 125$

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- At 2 hours, the traveler is 725 miles from home. --> TRUE. In fact, if we substitute t = 2 into the 1st equation,

$D(2) = (300)(2)+125 = 725$

- At 2.5 hours, the traveler is 875 miles from home. --> FALSE. In fact, for t=2.5 we have to use the 2nd equation, which states that the distance is:

D(t) = 880

So, not 875 miles.

- At 3 hours, the distance is constant, at 880 miles. --> TRUE. This is clearly visible from the 2nd equation: for t between 2.5 and 3.5 (so, in this case), the distance is

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$D(6)) = 75(6)+612.5=1062.5$