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Alona [7]
2 years ago
8

Hexagon D and hexagon H are regular hexagons. The scale factor from hexagon D to hexagon H is 0.25. One side of hexagon D measur

es 18 cm. What is the length of one side of hexagon H?
Mathematics
2 answers:
Alexus [3.1K]2 years ago
7 0
D/H = 0.25 18/H = 0.25/1 Cross multiply and solve for H
Charra [1.4K]2 years ago
7 0

Answer:

the answer is 4.5 cm

Step-by-step explanation:

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Solve each inequality. Graph the solution. *you must draw your graph on your work p-
V125BC [204]
54/2 or 27 should be the answer
3 0
3 years ago
2x - 2 = x-5<br> I’ll give 35 points plz help
tino4ka555 [31]

Answer:

x = -3

Step-by-step explanation:

2x-x=-5+2

x = -3

8 0
3 years ago
In the data set below, which of these points is most likely and outlier?
shusha [124]
<h2>Answer:</h2>

Option: D is the correct answer.

                 D.  (2,54)

<h2>Step-by-step explanation:</h2>

We know that an outlier of a data set is the value that stands out of the rest of the data point i.e. either it is a too high value or a too low value as compared to other data points.

Here  we are given a set of data points as:

         (2,54)

          (4,7)

         (6, 9)

         (8,12)

         (10,15)

Hence, we see that the output values i.e. 7 in (4,7) ; 9 in (6,9) ; 12 in (8,12) and 15 in (10,15) are closely related.

Hence, the data point that is an outlier is:

                  (2,54)

(As 54 is a much high value as compared to other)

5 0
3 years ago
The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is meas
saw5 [17]

Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by s(t) = -t^3+3t^2+7t +4, where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

s(t) = -t^3+3t^2+7t +4

Derivate w.r.t. time,

v(t)=s'(t) = -3t^2+6t+7

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

x=-\frac{b}{2a}

Where, a=-3, b=6 and c=7

t=-\frac{6}{2(-3)}

t=-\frac{6}{-6}

t=1

As 1 lie between interval [0,4]

Substitute t=1 in the function,

v(t)= -3(1)^2+6(1)+7

v(t)= -3+6+7

v(1)=10

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

8 0
2 years ago
Read 2 more answers
PLS HELP (CLICK ON HERE TO SEE QUESTION)
Vaselesa [24]

Answer:

angle y is 160 degrees

6 0
2 years ago
Read 2 more answers
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