All categories

3 years ago

Warm-Up: Perpendicular Bisector 1. What is WY? Explain your reasoning.

Mathematics

2 answers:

Lemur [1.5K]3 years ago

6 0

Answer: WY = 27

Step-by-step explanation:

Due to WX being a Perpendicular Bisector WY = WZ

4x - 5 = 2x + 11

Add 5 to both sides

4x = 2x + 16

Subtract 2x from both sides

2x = 16

Divide by 2

x = 8

So 4x - 5 of x = 8

4*8 = 32 - 5 =27

timofeeve [1]3 years ago

3 0

Answer:

Step-by-step explanation:

WY = WZ because XZ = XY

$2x + 11 = 4x - 5\\(2x + 11) - 11 = (4x - 5) -11\\2x = 4x - 16\\-2x = -16\\x = 8\\\\WY = 4x - 5\\WY = 4(8) -5\\WY = 32 - 5\\WY = 27$

You might be interested in

ANSWER QUICK I ONLY HAVE A WEEK BOEFORE I FAIL PLEASE ANSWER PIC IS THE QUESTION

deff fn [24]

Answer:

theres no picture

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Goran is riding his bicycle. He rides for 22.5 kilometers at a speed of 9 kilometers per hour. For how many hours does he ride?

astra-53 [7]

Answer: He rides for 2.5 hours or 2 and a half hours.

Step-by-step explanation:

He needs to complete a trajectory of 22.5 kilometers.

He rides 9 kilometers per hour.

Divide the total by the amount of kilometers that he rides.

$\frac{22.5}{9}=2.5h$

6 0

3 years ago

HELP PLEASE!! this is 8th grade math btw

natima [27]

Answer:

the 2nd one

Step-by-step explanation:

5 0

3 years ago

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0

4 years ago

Can someone please answer B (PICTURE)

Delicious77 [7]

Answer:

The question is asking for the perimeter. There are three ways to answer this depending on the assumptions though so I'll let you choose.

Assumptions are basically the different scenarios

1st scenario being the trim covers the carpet and the tile, they'd need 48m of trim (12+12+4+8+8+4)

2nd scenario in which the trim covers the carpet but not the tiled area, they'd need 44m of trim (12+12+4+8)

3rd scenario where the trim only covers the the tiled area, 4m of trim

(just 4 since the tile only touches 4m of the wall)

Hope this helps!

Step-by-step explanation:

trim is just the perimeter since it covers the base of the wall

7 0

3 years ago