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ivolga24 [154]
2 years ago
5

HELP!!

Mathematics
1 answer:
Liono4ka [1.6K]2 years ago
6 0

Answer:

Part 1: The polygon ABCDE reflected across y-axis to get the polygon MNOPQ. So, the polygon ABCDE congruent to polygon MNOPQ.

Part 2: The vertices of polygon VWXYZ are V(1, 3), W(-1, 7), X(-7, 7), Y(-5, 3), and Z(-3, 1).

Step-by-step explanation:

Part 1:

The vertices of the polygon ABCDE are A(2, 8), B(4, 12), C(10, 12), D(8, 8), and E(6, 6).

The vertices of the polygon MNOPQ are M(-2, 8), N(-4, 12), O(-10, 12), P(-8, 8), and Q(-6, 6).

We need to find the transformation or sequence of transformations that can be performed on polygon ABCDE to show that it is congruent to polygon MNOPQ.

The relation between the vertices of ABCDE and MNOPQ are defined as

(x,y)\rightarrow (-x,y)

It means the polygon ABCDE reflected across y-axis to get the polygon MNOPQ. So, the polygon ABCDE congruent to polygon MNOPQ.

Part 2:

If polygon MNOPQ is translated 3 units right and 5 units down, then

(x,y)\rightarrow (x+3,y-5)

M(-2,8)\rightarrow V(-2+3,8-5)=V(1,3)

N(-4,12)\rightarrow W(-4+3,12-5)=W(-1,7)

O(-10,12)\rightarrow X(-10+3,12-5)=X(-7,7)

P(-8,8)\rightarrow Y(-8+3,8-5)=Y(-5,3)

Q(-6,6)\rightarrow Z(-6+3,6-5)=Z(-3,1)

Therefore the vertices of polygon VWXYZ are V(1, 3), W(-1, 7), X(-7, 7), Y(-5, 3), and Z(-3, 1).

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A mapping diagram showing a relation, using arrows, between input and output for the following ordered pairs: (negative 3, negat
il63 [147K]

Answer:

  {x | x = –5, –3, 1, 2, 6}

Step-by-step explanation:

The domain is the list of first-values of the ordered pairs:

   {x | x = –5, –3, 1, 2, 6}

7 0
2 years ago
Read 2 more answers
The area of ABED is 49 square units. Given AGequals9 units and ACequals10 ​units, what fraction of the area of ACIG is represent
Mrac [35]

Answer:

The fraction of the area of ACIG represented by the shaped region is 7/18

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

In the square ABED find the length side of the square

we know that

AB=BE=ED=AD

The area of s square is

A=b^{2}

where b is the length side of the square

we have

A=49\ units^2

substitute

49=b^{2}

b=7\ units

therefore

AB=BE=ED=AD=7\ units

step 2

Find the area of ACIG

The area of rectangle ACIG is equal to

A=(AC)(AG)

substitute the given values

A=(9)(10)=90\ units^2

step 3

Find the area of shaded rectangle DEHG

The area of rectangle DEHG is equal to

A=(DE)(DG)

we have DE=7\ units

DG=AG-AD=9-7=2\ units

substituteA=(7)(2)=14\ units^2

step 4

Find the area of shaded rectangle BCFE

The area of rectangle BCFE is equal to

A=(EF)(CF)

we have

EF=AC-AB=10-7=3\ units

CF=BE=7\ units

substitute

A=(3)(7)=21\ units^2

step 5

sum the shaded areas

14+21=35\ units^2

step 6

Divide the area of  of the shaded region by the area of ACIG

\frac{35}{90}

Simplify

Divide by 5 both numerator and denominator

\frac{7}{18}

therefore

The fraction of the area of ACIG represented by the shaped region is 7/18

5 0
2 years ago
Circle C has radius of 10 cm. Each of points B and D is on the midpoint of the radius. Find the area of the shaded region.
olya-2409 [2.1K]
Simple :)
Area of shaded part = are of 1/4 circle - area of both triangles.
Are of circle = pie r^2 so 100x3.14 = 314 cm2.
Area of triangle AOB= area of triangleDOE = bh/2= 5x10/2= 25 each
However, the traingles share a common area which is quad DOB(I)
Lets take traingle AOE, whose are is bh/2=10x10/2=50cm2.
50-area of triangle A(I)E= 50-(
7 0
2 years ago
Read 2 more answers
Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 kg. and standard deviation 1.9 kg. Let X b
horrorfan [7]

Answer:

a. X\sim N(\mu = 6.1, \sigma = 1.9) b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349

Step-by-step explanation:

a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.

b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.

c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842

d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166

e. The z-score related to 6.4 kg is z_{1} = (6.4-6.1)/1.9 = 0.1579 and the z-score related to 7 kg is z_{2} = (7-6.1)/1.9 = 0.4737, we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194

f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349

7 0
2 years ago
The price of a ring was increased by 9% to £1800. What was the price before the increase? Give your answer to the nearest penny.
tekilochka [14]

Answer:

1651

Step-by-step explanation:

let s say that the price before the increase is x

to apply an increase of 9% it does x + x*0.09 = x*(1+0.09)=x*1.09

and we know that this value is 1800

so

x*1.09=1800

<=>

x = 1800/1.09=1651.376147

to the nearest penny it gives 1651

4 0
2 years ago
Read 2 more answers
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