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ra1l [238]
2 years ago
12

Amira is writing a coordinate proof to show that the area of a triangle created by joining the midpoints of an isosceles triangl

es is one-fourth the area of the isosceles triangle. She starts by assigning coordinates as given.
Triangle D E F in the coordinate plane so that vertex D is at the origin and is labeled 0 comma 0, vertex E is in the first quadrant and is labeled 2 a comma 2 b, and vertex F is on the positive side of the x-axis and is labeled 4a comma 0. Point Q is between points D and E. Point R is between points E and F. Point P is between points D and F and is labeled 2 a comma 0.

Enter your answers, in simplest form, in the boxes to complete the coordinate proof.
Point Q is the midpoint of DE¯¯¯¯¯ , so the coordinates of point Q are (a, b) .

Point R is the midpoint of FE¯¯¯¯¯ , so the coordinates of point R are (
, b).

In △DEF , the length of the base, DF¯¯¯¯¯ , is
, and the height is 2b, so its area is
.

In △QRP , the length of the base, QR¯¯¯¯¯ , is
, and the height is b, so its area is ab .

Comparing the expressions for the areas proves that the area of the triangle created by joining the midpoints of an isosceles triangle is one-fourth the area of the

Mathematics
2 answers:
vodomira [7]2 years ago
7 0

Answer: We can fill the boxes with help of below explanation.

Explanation: According to the given figure,

Triangle DEF is an isosceles triangle with sides DE, EF and FD.

And, where  DE=EF and P,R and Q are the midpoints of the sides DF, EF and DE respectively.

Theorem: Area of a triangle created by joining the midpoints of an isosceles triangles is one-fourth the area of the isosceles triangle.

Proof:

Point Q is the midpoint of DE, so the coordinates of point Q are (a, b). (because mid point of a line segment always has coordinates half of the sum of corresponding coordinates of end points. )

Point R is the midpoint of FE , so the coordinates of point R are (

3a,b).

In triangle DEF , the length of the base, DF is  4a and the height is 2b, So, its area is 4ab.( because, Area of a triangle= 1/2×base×height

In triangle QRP , the length of the base, QR is   2a and the height is b,

So, its area is ab.

On comparing the above two expression, we found that, area of triangle DEF =4( area of triangle QRP)





Romashka [77]2 years ago
6 0
Point R is the midpoint of FE¯¯¯¯¯ , so the coordinates of point R are (3a, b).

In △DEF , the length of the base, DF¯¯¯¯¯ , is
4a, and the height is 2b, so its area is
1/2×4a×2b = 4ab.

In △QRP , the length of the base, QR¯¯¯¯¯ , is
3a-a = 2a, and the height is b, so its area is 1/2×2a×b = ab .

Comparing the expressions for the areas proves that the area of the triangle created by joining the midpoints of an isosceles triangle is one-fourth the area of the larger isosceles triangle.
You might be interested in
Jonathan borrowed $475 at a simple annual interest rate of 2%. How many years will it take him to repay the loan if he wants to
vovikov84 [41]
Firstly you would times 475 by the percentage, which you can convert to a decimal my multiplying the percentage by 100.

475 * 0.02 = 9.50

Then divide the interest by the amount of interest per year:
38 / 9.5 = 4.

The answer would be 4 years.

Hope this helps.
3 0
2 years ago
Read 2 more answers
The Big River Casino is advertising a new digital lottery-style game called Instant Lotto. The player can win the following mone
zysi [14]

Answer:

(a) The expected value of the prize for one play of Instant Lotto is $3.50.

(b) The probability that the visitor wins some prize at least twice in the 20 free plays is 0.2641.

(c) The probability that a randomly selected day has at least 1000 people play Instant Lotto is 0.2579.

Step-by-step explanation:

(a)

The probability distribution of the monetary prizes that can be won at the game called Instant Lotto is:

<em>X</em>         P (<em>X</em> = <em>x</em>)

$10        0.05

$15        0.04

$30       0.03

$50       0.01

$1000   0.001

$0         0.869

___________

Total =   1.000

Compute the expected value of the prize for one play of Instant Lotto as follows:

E(X)=\sum x\cdot P (X=x)

         =(10\times 0.05)+(15\times 0.04)+(30\times 0.03) \\+ (50\times 0.01)+(1000\times 0.001)+(0\times 0.869)\\=0.5+0.6+0.9+0.5+1+0\\=3.5          

Thus, the expected value of the prize for one play of Instant Lotto is $3.50.

(b)

Let <em>X</em> = number of times a visitor wins some prize.

A visitor to the casino is given <em>n</em> = 20 free plays of Instant Lotto.

The probability that a visitor wins at any of the 20 free plays is, <em>p</em> = 1/20 = 0.05.

The event of a visitor winning at a random free play is independent of the others.

The random variable <em>X</em> follows Binomial distribution with parameters <em>n</em> = 20 and <em>p</em> = 0.05.

Compute the probability that the visitor wins some prize at least twice in the 20 free plays as follows:

P (X ≥ 2) = 1 - P (X < 2)

              = 1 - P (X = 0) - P (X = 1)

              =1-[{20\choose 0}0.05^{0}(1-0.05)^{20-0}]-[{20\choose 1}0.05^{1}(1-0.05)^{20-1}]\\=1-0.3585-0.3774\\=0.2641

Thus, the probability that the visitor wins some prize at least twice in the 20 free plays is 0.2641.

(c)

Let <em>X</em> = number of people who play Instant Lotto each day.

The random variable <em>X</em> is normally distributed with a mean, <em>μ</em> = 800 people and a standard deviation, <em>μ</em> = 310 people.

Compute the probability that a randomly selected day has at least 1000 people play Instant Lotto as follows:

Apply continuity correction:

P (X ≥ 1000) = P (X > 1000 + 0.50)

                    = P (X > 1000.50)

                    =P(\frac{X-\mu}{\sigma}>\frac{1000.50-800}{310})

                    =P(Z>0.65)\\=1-P(Z

Thus, the probability that a randomly selected day has at least 1000 people play Instant Lotto is 0.2579.

5 0
2 years ago
Milton is floating in an inner tube in a wave pool. He is 1.5 m from the bottom of the pool when he is at the trough of a wave.
posledela
This is the concept of sinusoidal, to solve the question we proceed as follows;
Using the formula;
g(t)=offset+A*sin[(2πt)/T+Delay]
From sinusoidal theory, the time from trough to crest is normally half the period of the wave form. Such that T=2.5
The pick magnitude is given by:
Trough-Crest=
2.1-1.5=0.6 m
amplitude=1/2(Trough-Crest)
=1/2*0.6
=0.3
The offset to the center of the circle is 0.3+1.5=1.8
Since the delay is at -π/2 the wave will start at the trough at [time,t=0]
substituting the above in our formula we get:
g(t)=1.8+(0.3)sin[(2*π*t)/2.5]-π/2]
g(t)=1.8+0.3sin[(0.8πt)/T-π/2]


3 0
2 years ago
Find the distance from (4, −7, 6) to each of the following.
LenKa [72]

Answer:

(a) 6 units

(b) 4 units

(c) 7 units

(d) 9.22 units

(e) 7.21 units

(f) 8.06 units

Step-by-step explanation:

The distance d from one point (x₁, y₁, z₁) to another point (x₂, y₂, z₂) is given by;

d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

Now from the question;

<em>(a) The distance from (4, -7, 6) to the xy-plane</em>

The xy-plane is the point where z is 0. i.e

xy-plane = (4, -7, 0).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, -7, 0)</em>

d = √[(4 - 4)² + (-7 - (-7))² + (0 - 6)²]

d = √[(0)² + (0)² + (-6)²]

d = √(-6)²

d = √36

d = 6

Hence, the distance to the xy plane is 6 units

<em>(b) The distance from (4, -7, 6) to the yz-plane</em>

The yz-plane is the point where x is 0. i.e

yz-plane = (0, -7, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, -7, 6)</em>

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 6)²]

d = √[(4)² + (0)² + (0)²]

d = √(4)²

d = √16

d = 4

Hence, the distance to the yz plane is 4 units

<em>(c) The distance from (4, -7, 6) to the xz-plane</em>

The xz-plane is the point where y is 0. i.e

xz-plane = (4, 0, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, 0, 6)</em>

d = √[(4 - 4)² + (-7 - 0)² + (6 - 6)²]

d = √[(0)² + (-7)² + (0)²]

d = √[(-7)²]

d = √49

d = 7

Hence, the distance to the xz plane is 7 units

<em>(d) The distance from (4, -7, 6) to the x axis</em>

The x axis is the point where y and z are 0. i.e

x-axis = (4, 0, 0).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, 0, 0)</em>

d = √[(4 - 4)² + (-7 - 0)² + (6 - 0)²]

d = √[(0)² + (-7)² + (6)²]

d = √[(-7)² + (6)²]

d = √[(49 + 36)]

d = √(85)

d = 9.22

Hence, the distance to the x axis is 9.22 units

<em>(e) The distance from (4, -7, 6) to the y axis</em>

The x axis is the point where x and z are 0. i.e

y-axis = (0, -7, 0).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, -7, 0)</em>

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 0)²]

d = √[(4)² + (0)² + (6)²]

d = √[(4)² + (6)²]

d = √[(16 + 36)]

d = √(52)

d = 7.22

Hence, the distance to the y axis is 7.21 units

<em>(f) The distance from (4, -7, 6) to the z axis</em>

The z axis is the point where x and y are 0. i.e

z-axis = (0, 0, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, 0 6)</em>

d = √[(4 - 0)² + (-7 - (0))² + (6 - 6)²]

d = √[(4)² + (-7)² + (0)²]

d = √[(4)² + (-7)²]

d = √[(16 + 49)]

d = √(65)

d = 8.06

Hence, the distance to the z axis is 8.06 units

5 0
2 years ago
A law firm is going to designate associates and partners to a big new case. The daily rate charged to the client for each associ
OleMash [197]

Answer:

Each associate costs the client $700 per day, so a associates will cost the client 700a dollars per day. Each partner costs the client $1500 per day, so p partners will cost the client 1500p dollars per day. The total amount charged 700a+1500p :$14100:

700a+1500p=14100

700a+1500p=14100

Since there were a total of 11 lawyers assigned to the case (associates and partners), we know a+pa+p must equal 11.11.

a+p=11

a+p=11

\underline{\text{Write System of Equations:}}

Write System of Equations:

​

700a+1500p=

700a+1500p=14100

14100

a+p=11

11

8 0
2 years ago
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