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vladimir2022 [97]
2 years ago
10

Y= -7 and -4 + 2y = -2

Mathematics
1 answer:
IgorLugansk [536]2 years ago
5 0

Answer:

Hope this helps! :)

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What is the area of the parallelogram?<br> 48 sqrt3 cm^2<br> 48 cm^2<br> 24 sqrt3 cm^2<br> 24 cm^2
Salsk061 [2.6K]
The area of a parallelogram is base times height. 

We can find the height by deriving a right triangle from the 60 degree angle and the 6 cm hypotenuse

sin(60°) = opp/hyp = x/6
√3/2 = x/6     Multiply both sides by 6
3√3 = x
x = 3√3

The height is 3 √3

Area = base x height
Area = 3√3 * 8
Area = 24 √3 cm²

Answer is C) 24 √3 cm²

3 0
3 years ago
A 25-foot long ladder is propped against a wall at an angle of 18° with the wall. how high up the wall does the ladder reach? ro
Alecsey [184]

Answer

Find out the how high up the wall does the ladder reach .

To proof

let us assume that the height of the wall be x .

As given

A 25-foot long ladder is propped against a wall at an angle of 18° .

as shown in the diagram given below

By using the trignometric identity

cos\theta = \frac{Base}{Hypotenuse}

now

\theta = 18^{\circ}

Base = wall height = x

Hypotenuse = 25 foot

Put in the trignometric identity

cos18^{\circ} = \frac{x}{25}

cos18^{\circ} = \bar{0.9510565162}

x = 23.8 foot ( approx)

Therefore the height of the  ladder be 23.8 foot ( approx) .

8 0
3 years ago
Read 2 more answers
HELPPP! Last attempt geometry!!
Helen [10]

Answer:

  x1 = 2, y1 = 0

  x2 = 2, y2 = 6

Step-by-step explanation:

Each pair of coordinates is (x, y). The first pair is (x1, y1). The second pair is (x2, y2). This means you have ...

  x1 = 2, y1 = 0

  x2 = 2, y2 = 6

_____

<em>Additional comment</em>

The slope of the line through these points is ...

  m = (y2 -y1)/(x2 -x1) = (6 -0)/(2 -2) = 6/0 = undefined

The line through these points is a vertical line with equation x = 2.

6 0
3 years ago
Put these numbers in order from least to greatest: -4 2/5 -4.5 -5.1 -4.45 -5 -
liberstina [14]

Answer:

-5.1, -5, -4.5, -4.45,-4 2/5

3 0
3 years ago
Read 2 more answers
Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL
Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

  • J(2;5); K(1;1)

             => JK = \sqrt{(1-2)^{2} + (1-5)^{2}  } = \sqrt{(-1)^{2}+(-4)^{2}  } = \sqrt{1+16}=\sqrt{17}

  • J(2;5); L(5;2)

             => JL = \sqrt{(5-2)^{2} + (2-5)^{2}  } = \sqrt{3^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • K(1;1); L(5;2)

             =>  KL = \sqrt{(5-1)^{2} + (2-1)^{2}  } = \sqrt{4^{2}+1^{2}  } = \sqrt{1+16}=\sqrt{17}

In the triangle QNP, the sides can be calculate as following:

  • Q(-4;4); N(-3;0)

             => QN = \sqrt{[-3-(-4)]^{2} + (0-4)^{2}  } = \sqrt{1^{2}+(-4)^{2}  } = \sqrt{1+16}=\sqrt{17}

  • Q (-4;4); P(-7;1)

   => QP = \sqrt{[-7-(-4)]^{2} + (1-4)^{2}  } = \sqrt{(-3)^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • N(-3;0); P(-7;1)

             =>  NP = \sqrt{[-7-(-3)]^{2} + (1-0)^{2}  } = \sqrt{(-4)^{2}+1^{2}  } = \sqrt{16+1}=\sqrt{17}

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

7 0
3 years ago
Read 2 more answers
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