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tatyana61 [14]
3 years ago
5

A polygon is transformed using the rule (x, y) (3x, 3y). Which statement best describes the transformation?

Mathematics
2 answers:
sleet_krkn [62]3 years ago
5 0
Ummmmmm I think is A
FinnZ [79.3K]3 years ago
5 0

Answer:

<h3>A) an enlargement by scale factor of 3</h3>

Step-by-step explanation:

You might be interested in
Quadrilateral ABCD is located at A (−2, 2), B (−2, 4), C (2, 4), and D (2, 2). The quadrilateral is then transformed using the r
Mademuasel [1]
New cordinates are formed by adding 7 in x and subtracting 2 from y
A(−2, 2) =A ' (-2 +7 , 2 - 1 ) = A' (5,1)
B(−2, 4)  = B' (-2 + 7 , 4 -1 )= B' (5,3)
C(2, 4)   =  C' (2 + 7 , 4 -1 )= C' (9,3)
<span>D(2, 2)   D' (2 + 7 , 2 -1 ) = D' ( 9 , 1)</span><span>

</span>
8 0
3 years ago
Jada solved the equation Negative StartFraction 4 over 9 EndFraction = StartFraction x over 108 EndFraction for x using the step
GalinKa [24]

Answer:

Jada should have multiplied both sides of the equation by 108.

Step-by-step explanation:

The question is incomplete. Find the complete question in the comment section.

Given the equation -4/9 = x/108, in order to determine Jada's error, we need to solve in our own way as shown:

Step 1: Multiply both sides of the equation by -9/4 as shown:

-4/9 × -9/4 = x/108 × -9/4

-36/-36 = -9x/432

1 = -9x/432

1 = -x/48

Cross multiplying

48 = -x

x = -48

It can also be solved like this:

Given -4/9 = x/108

Multiply both sides by 108 to have:

-4/9 * 108 = x/108 * 108

-4/9 * 108 = 108x/108

-432/9 = x

x = -48

Jada should have simply follow the second calculation by multiplying both sides of the equation by 108 as shown.

7 0
3 years ago
Kelly runs a restaurant that sells two kinds of desserts. Kelly knows the restaurant must make at least 25 dozen and at most 45
MissTica

Answer:

  (c, m) = (45, 10)

Step-by-step explanation:

A dozen White Chocolate Blizzards generate more income and take less flour than a dozen Mint Breezes, so production of those should clearly be maximized. Making 45 dozen Blizzards does not use all the flour, so the remaining flour can be used to make Breezes.

Maximum Blizzards that can be made: 45 dz. Flour used: 45×5 oz = 225 oz.

The remaining flour is ...

  315 oz -225 oz = 90 oz

This is enough for (90 oz)/(9 oz/dz) = 10 dozen Mint Breezes. This is in the required range of 2 to 15 dozen.

Kelly should make 45 dozen White Chocolate Blizzards and 10 dozen Mint Breezes: (c, m) = (45, 10).

__

In the attached graph, we have reversed the applicable inequalities so the feasible region shows up white, instead of shaded with 5 different colors. The objective function is the green line, shown at the point that maximizes income. (c, m) ⇔ (x, y)

8 0
3 years ago
Find the area of the shape shown below
Dovator [93]

Answer:

48

Step-by-step explanation:

the answer is 48

7 0
3 years ago
Read 2 more answers
The following data represent the monthly phone use, in minutes, of a customer enrolled in a fraud prevention program for the pas
Illusion [34]

Answer:

The answer is "717.25 minutes".

Step-by-step explanation:

In this question first, we arrange the value in ascending order that are:

307,311,321,322,354,363,377,406,425,435,461,464,474,499,505,513,517,534,548

The median is always in an orderly position that is = \frac{(n+1)}{2}.

\to \frac{(n+1)}{2} = \frac{(20+1)}{2} = 10.5  position orders

10^{th} \ and \ 11^{th} place an average of observations so, the

Average = \frac{(425 +435)}{2} = \frac{(860)}{2} =430

Because as medium stands at 10.5 that median is as below 10 are greater than the level.

Q_1 often falls throughout the ordered BELOW average is = \frac{(n+1)}{2} place.

\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th}ordered position

Average 5^{th} \ and \ 6^{th}  location findings

Q_1 = \frac{(354+363)}{2} = \frac{(717)}{2}= 358.5

In  = \frac{(n+1)}{2} the ordered place, Q3 always falls ABOVE the median.

\to \frac{(n+1)}{2} = \frac{(10+1)}{2} = 5.5^{th} ordered At median ABOVE 

Consequently Q_3 falls between 5^{th} \ and \ 6^{th} ABOVE the median position

15^{th}\  and \ 16^{th}place average of observations

\to Q_3 = \frac{(499+505)}{2}=\frac{(1004)}{2}  = 502

\to IQR = Q_3 - Q_1= 502-358.5= 143.5  

\to \text{Upper fence} = Q_3 + 1.5 \times IQR= 502 + 1.5 \times 143.5 = 717.25

4 0
3 years ago
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