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Elenna [48]
3 years ago
7

Which classification best describes ∠2?

Mathematics
2 answers:
Salsk061 [2.6K]3 years ago
8 0
The answer is C) obtuse
soldier1979 [14.2K]3 years ago
8 0

Answer:

C. Obtuse.

Step-by-step explanation:

We are asked to classify the type of angle 2.

We can see that measure of angle 2 is neither a straight line nor a right angle.

We can also see that measure of angle 2 is greater than 90 degree, therefore, angle 2 is not an acute angle.

Since the measure of angle 2 is greater than 90 degrees, therefore, angle 2 is an obtuse angle.

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If x=-1, x^5+x+4+x^3+x
DedPeter [7]

Answer:

0

Step-by-step explanation:

yeah-ya.............. right?

8 0
2 years ago
Read 2 more answers
Can someone plz help me solved this problem I need help plz help me! Will mark you as brainiest!
I am Lyosha [343]

Answer:

13 l of 15%, 13 l of 25%, 26 l of 70%

Step-by-step explanation:

solutions:

  • 15%, 25% and 70%

final solution:

  • 52 l of 45%

This includes 0.45*52l= 23.4 l acid

Let's assume:

  • amount of 15% solution= x
  • amount of 25% solution= y
  • amount of 70% solution= z
  • and we have z=2y

so there are 2 equation:

  • x+y+z=52
  • 0.15x+0.25y+0.7z=23.4

or

  • x+y+2y= 52 ⇒ x+3y= 52 ⇒ x= 52- 3y
  • 0.15x+0.25y+1.4y=23.4
  • 0.15x+1.65y=23.4
  • 0.15*(52-3y)+1.65y=23.4
  • 7.8-0.45y+1.65y=23.4
  • 1.2y=23.4-7.8
  • 1.2y= 15.6
  • y=15.6/1.2
  • y= 13 l

--------

  • x= 52- 3*13= 13 l
  • z= 2y= 2*13= 26 l
6 0
3 years ago
The coordinates of polygon ABCD are A(-4,-1), B(-2,3), C(2,2), and D(4,-3). Use these coordinates to complete the sentences belo
oee [108]

Answer:

The perimeter of polygon ABCD, to the nearest thousandth units is

22.227 units

The perimeter of polygon ABCD', to the nearest thousandth, would be 20.980 units

The area of polygon ABCD' is 19.5 units²

Step-by-step explanation:

The coordinates forming the polygon are

A (-4,-1), B(-2,3), C(2,2), and D(4,-3)

The perimeter then is given by the sum of the length of the sides as follows;

Length of line between two X and Y points distance, xi and yj apart is

length XY =  \sqrt{x^2+y^2}

Therefore, the length between points AB is

length AB = \sqrt{(-4 - (-2))^2+(-1-3)^2}

= \sqrt{(-4 +2)^2+(-4)^2} = \sqrt{-2^2+-4^2}  = \sqrt{20}  = 4.472 units

Similarly, length BC is given by

Length BC =  \sqrt{(-4)^2+1^2} =  \sqrt{17} = 4.123 units

Length CD = \sqrt{(-2)^2+5^2} =  \sqrt{29} = 5.385 units

Length DA = \sqrt{(8)^2+(-2)^2} =  \sqrt{68} = 8.246 units

The perimeter is equal to;

length AB + Length BC + Length CD + Length DA

= 4.472 + 4.123 + 5.385 + 8.246 = 22.227 units

If the point D is moved up 2 units and left 1 unit we have

D' = x-1, y+2 where x and y are the coordinates of point D

D(4, -3) → D'((4-1), (-3+2)) = D'(3, -1)

The length D'A =  \sqrt{(7)^2+(0)^2} =  \sqrt{49} =7 units

The perimeter of polygon ABCD'

length AB + Length BC + Length CD + Length D'A

= 4.472 + 4.123 + 5.385 + 7 = 20.980 units.

The area is given by the determinant of the 3 by 3 matrix using Cramer's Rule as follows

 S_{\bigtriangleup} = (\frac{1}{2})|x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2|

= (\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|

Triangle ABC we have

A(-4,-1)

B(-2,3)

C(2,2)

S_{{\bigtriangleup}ABC} = (\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|

=(\frac{1}{2})|(-4)(3-2)+(-2)(2-(-1)) +2((-1)-3)|

= =(\frac{1}{2})|-4-6 -8|= 9 units^2

Triangle ACD'

A(-4,-1)

C(2,2)

D'(3, -1)

S_{{\bigtriangleup}ACD'} = (\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|

=(\frac{1}{2})|(-4)(2+1)+(2)(-1+1) +3((-1)-2)| = \frac{21}{2} = 10.5 units²

Area of polygon =   S_{{\bigtriangleup}ABCD'} = S_{{\bigtriangleup}ABC} + S_{{\bigtriangleup}ACD'} = (9 + 10.5) units²

Area of polygon ABCD' = 19.5 units².

7 0
3 years ago
What is the Mean medium and mode 1 2 2 2 3 4 4
Sindrei [870]
Mean ; 2.6
Median; 2
Mode ; 2
8 0
3 years ago
Read 2 more answers
What is 10t =90 and how do I solve this?please help
Nesterboy [21]
To solve this you have to divide 90 by 10 which would be 9 so


t=9

Hope it helps
6 0
4 years ago
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