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ivanzaharov [21]
3 years ago
12

Worth a  brainliest if the answer is correct.

Mathematics
1 answer:
ycow [4]3 years ago
6 0
You need to translate the angles using this function (x,y) to (x+3,y-3)
first of all find A and B and C
A=(0,3),B=(-1,1),C=(2,1)
After the translation
A'=(3,0),B'=(2,-2),C'=(5,-2)
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Solve for x Enter the solution from least to greatest (x+6)(-x+1)=0
Sunny_sXe [5.5K]

Answer:

Step-by-step explanation:

x + 6 = 0

x = -6

-x + 1 = 0

-x = -1

x = 1

x = -6, 1

3 0
3 years ago
Plz help me with this question
Mila [183]
The correct answer is 1 inch.
4 0
3 years ago
Read 2 more answers
Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3> (1 point)
sesenic [268]

Angle between u = -5i-4j , v=-4i-3j is x =0° .

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

We have , two vectors u = <-5, -4>, v = <-4, -3>  or , u = -5i-4j , v=-4i-3j

We need to find angle between these two vectors . Let's find out:

We know that dot product of two vectors is defined as :

u.v =|u|(|v|)cosx , where x is angle between u & v !

⇒ u.v =|u|(|v|)cosx

⇒ cosx =\frac{u.v}{|u|(|v|)}

Now , u.v = (-5i-4j)(-4i-3j)

⇒ u.v = (-5i-4j)(-4i)-(-5i-4j)(3j)

⇒ u.v = 20+12            { i(j) = j(i) =0  }

⇒ u.v = 32

Now , Modulus of any vector  r = xi+yj is |r| = \sqrt{x^{2}+y^{2}} So ,

|u| = \sqrt{(-5)^{2}+(-4)^{2}} = \sqrt{25+16} = \sqrt{41} \\\\|v| = \sqrt{(-4)^{2}+(-3)^{2}} = \sqrt{16+9} = \sqrt{25} = 5

Putting all these values in equation cosx =\frac{u.v}{|u|(|v|)} we get:

⇒ cosx =\frac{32}{5(\sqrt{41})}

⇒ cos^{-1}(cosx) =cos^{-1}(\frac{32}{5(6.4)})

⇒ x =cos^{-1}(1)                 { cos0 = 1  }

⇒ x =0°

Therefore , Angle between u = -5i-4j , v=-4i-3j is x =0° .

8 0
3 years ago
What is 2 to the 0 power in expression form
horrorfan [7]

Answer:

1

Step-by-step explanation:

8 0
3 years ago
Helpppppppppppppppppppoppppppoo
postnew [5]

Answer:

Step-by-step explanation:

abc=80, abm= 100

acb= 40, acn= 140

angle a = 60

x=80, y=40

8 0
3 years ago
Read 2 more answers
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