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madam [21]
1 year ago
5

the vertices of abc are a (-6,-7) b(-3,-10) and c(-5,2) find the vertices of abc given the transition rule (x,y)-->(x,y-3)

Mathematics
1 answer:
bagirrra123 [75]1 year ago
4 0

The vertices of abc given the transition rule (x,y)-->(x,y-3) are (-6, -10), (-3, -13) and (-5, -1)

<h3>How to determine the new vertices?</h3>

The vertices are given as:

a (-6,-7)

b(-3,-10)

c(-5,2)

The transition rule is given as

(x,y)-->(x,y-3)

So, we have

a' = (-6, -7 - 3)

a' = (-6, -10)

b' = (-3, -10 - 3)

b' = (-3, -13)

c' = (-5, 2 - 3)

c' = (-5, -1)

Hence, the vertices of abc given the transition rule (x,y)-->(x,y-3) are (-6, -10), (-3, -13) and (-5, -1)

Read more about translation at:

brainly.com/question/4289712

#SPJ1

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ad-work [718]
\dfrac{\cos^3A-\cos A}{\sin^3A-\sin A}+\dfrac{\cos^2A-\cos^4A}{\sin^4A-\sin^2A}=\dfrac{\sin A}{\cos^2A\cos^3A}\\\\L_s=\dfrac{\cos A(\cos^2A-1)}{\sin A(\sin^2A-1)}+\dfrac{\cos^2A(1-\cos^2A)}{\sin^2A(\sin^2A-1)}\\\\=\dfrac{\cos A(-\sin^2A)}{\sin A(-\cos^2A)}+\dfrac{\cos^2A\sin^2A}{\sin^2A(-\cos^2A)}\\\\=\dfrac{\sin A}{\cos A}-1=\tan A-1

R_s=\dfrac{\sin A}{\cos A\cos^4A}=\dfrac{\sin A}{\cos A}\cdot\dfrac{1}{\cos^4A}=\tan A\cdot\dfrac{1}{\cos^4A}=\dfrac{\tan A}{\cos^4A}


\boxed{L_s\neq R_s}
8 0
3 years ago
Jim and Em baked 33 macaroons.
bearhunter [10]

Answer: 19 macaroons

Step-by-step explanation:

Let's create an equation to represent this situation. Let x represent the number of macaroons that Em made.

33=x+x+5

33=2x+5

33-5=2x+5-5

28=2x

28/2=2x/2

14=x

This means that Em made 14 macaroons.

14+5=19

Therefore, Jim baked 19 macaroons.

4 0
3 years ago
A software distributor charges \$41$41dollar sign, 41 per license of a particular utility. The distributor offers a 10\%10%10, p
Alchen [17]

Answer:

1558

Step-by-step explanation:

The first 202020 licenses cost full price, which is \$41$41dollar sign, 41 each. Thus, the first 202020 licenses cost a total of \$41 \cdot 20 = \$820$41⋅20=$820dollar sign, 41, dot, 20, equals, dollar sign, 820.

Hint #2

The remaining licenses have a 10\%10%10, percent discount. That means they cost 90\%90%90, percent of the full price. Let's write 90\%90%90, percent in its decimal form, 0.90.90, point, 9, in order to multiply.

\qquad 0.9 \cdot \$41 = \$36.900.9⋅$41=$36.900, point, 9, dot, dollar sign, 41, equals, dollar sign, 36, point, 90

Hint #3

There are 40 - 20 = 2040−20=2040, minus, 20, equals, 20 licenses at the reduced price of \$36.90$36.90dollar sign, 36, point, 90. Thus, the remaining 202020 licenses cost a total of \$36.90 \cdot 20 = \$738$36.90⋅20=$738dollar sign, 36, point, 90, dot, 20, equals, dollar sign, 738.

Hint #4

To find the cost for all 404040 licenses, let's add the costs of the full price and discounted licenses.

\qquad \$820 + \$738 = \$1558$820+$738=$1558dollar sign, 820, plus, dollar sign, 738, equals, dollar sign, 1558

Hint #5

The business would pay \$1558$1558dollar sign, 1558 to buy the licenses.

3 0
3 years ago
Please help
Ann [662]

Answer:

1/60 probabiliity

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You have two independent events that you want to put together.

Let Pr. mean "probability"

Pr(5 from 10 cards and 2 on a dice ) = Pr(5 from 10 cards) * Pr( 2 on a dice)

Pr(5 from 10 cards and 2 on a dice ) =(1/10) * (1/6)

= 1/60

Pr(5 from 10 cards and 2 on a dice ) =0.0167

or 1.67% probability

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