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Nesterboy [21]
2 years ago
3

The red square is the image of the blue square after a single transformation. Use two or more complete sentences to describe eac

h of the four different transformations that could produce the image.
Mathematics
2 answers:
Bond [772]2 years ago
8 0
The red square could have been dilated to fit the blue square. Or maybe one of the square's was reflected off of another using the x or y axis. Maybe the red square was rotated clockwise or counter-clockwise to fit the blue square. The last things is maybe the red square was translated a certain amount of units up or down to fit the blue square.   <span />
frez [133]2 years ago
5 0

The blue square can be rotated 180° or -180° to fit the red square.

The blue square can be reflected over y = -x to fit the red square.

The blue square can be translated to the left and then down <em>or</em> down and then to the left to fit the red square.

The blue square can be dilated by -1 to fit the red square.

<em>Hope this helps!</em>

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A circle with radius of \greenD{5\,\text{cm}}5cmstart color #1fab54, 5, start text, c, m, end text, end color #1fab54 sits insid
Marat540 [252]

Answer:

The area of the shaded region is 42.50 cm².

Step-by-step explanation:

Consider the figure below.

The radius of the circle is, <em>r</em> = 5 cm.

The sides of the rectangle are:

<em>l</em> = 11 cm

<em>b</em> = 11 cm.

Compute the area of the shaded region as follows:

Area of the shaded region = Area of rectangle - Area of circle

                                            =[l\times b]-[\pi r^{2}]\\\\=[11\times 11]-[3.14\times 5\times 5]\\\\=121-78.50\\\\=42.50

Thus, the area of the shaded region is 42.50 cm².

7 0
2 years ago
Read 2 more answers
The number of tickets issued by a meter reader for parking-meterviolations can be modeled by a Poisson process with a rateparame
Doss [256]

Answer:

Step-by-step explanation:

Let X be the  number of tickets issued by a meter reader for parking-meterviolations can be modeled by a Poisson process with a rateparameter of five per hour.

X is Poisson with parameter =5per hour

a) the probabilitythat exactly three tickets are given out during a particular hour

=P(X=3)=0.14037

b)  the probabilitythat at least three tickets are given out during a particularhour

=P(X\geq 3) =0.87534

c) tickets we expect to be given during a 45-min period

=5(\frac{3}{4} )\\= 3.75

Note: Poisson distribution is

P(X=k) = \frac{e^{-k} 5^k}{l!}

3 0
2 years ago
Show that the three points whose position vectors are 7j+ 10k,-i + 6j+6k and - 4i + +9j + 6k form an isosceles right
Novay_Z [31]

Answer:

AB = √18 , BC=√18 and CA =4

AB²+BC²  = CA² and AB=BC

ΔABC isosceles right  angled triangle.

Step-by-step explanation:

Given vectors are  7j+ 10k,-i + 6j+6k and - 4i + +9j + 6k

A( 0,7,10), B( -1,6,6) C(-4,9,6)

AB⁻ = OB-OA = -I+6j+6k-(7j+10k) = -I-j-4k

AB = \sqrt{1+1+16} = \sqrt{18}

BC = OC-OB = -4i+9j+6k-(-I+6j+6k) = -3i+3j

BC=\sqrt{9+9} =\sqrt{18}

CA = OA-OC = 7j+10k - (- 4i + +9j + 6k ) = 4i-2j+4k

CA = \sqrt{16+4+16} =\sqrt{36} =4

Since AB²+BC²  = CA²

And AB=BC

Therefore it follows that ΔABC is a right angled isosceles triangle



3 0
2 years ago
The exponential functions y=(1-25)^x-2/5. -10 is shown hraphed along woth the horizontal line y=115 their intersection is (a,115
evablogger [386]

Answer:26

Step-by-step explanation:

Y=1.25^x-2/5-10

Take log of both sides

So Y+10=1.25^x-2/5

So log to base 10 of the two sides of the equation is

Log(Y+10)=X-2/5log1.25.

To make X the subject, divide both sides by log1.25.

Log(Y+10)/log1.25=X-2/5.

Recall that Y was given to be 115

It becomes log(115 +10)/Log1.25=x-0.4

21.64=x-0.4

X=25.6

3 0
2 years ago
The table shows the concentration of a reactant in the reaction mixture over a period of time.
saveliy_v [14]
<h3>Option C</h3><h3>The average rate of the reaction over the entire course of the reaction is: 2.0 \times 10^{-3}</h3>

<em><u>Solution:</u></em>

Average rate is the ratio of concentration change to the time taken for the change

Average\ rate = \frac{ \triangle c }{\triangle t }

The concentration of the reactants changes 1.8 M to 0.6 M

here, the time interval given is 0 to 580 sec

Therefore,

Average\ rate = \frac{1.8-0.6}{580} \\\\Average\ rate = \frac{1.2}{580} \\\\Average\ rate \approx 2.0 \times 10^{-3}

Thus option C is correct

5 0
2 years ago
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