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BigorU [14]
3 years ago
11

I’m literally dumb pls help me

Mathematics
1 answer:
Verdich [7]3 years ago
7 0

Answer:

32.4 ft

Step-by-step explanation:

the perimeter of the half circle = ( 1/2 * 3.1 * 8) - 8 = 4.4 ft

the total perimeter = 10 + 8 +10 + 4.4 = 32.4 ft

( the right-side width of the rectangle is the diameter of the circle because it passes the origin point. Then you just need to calculate the perimeter of that half circle. Remebering to subtract the length of that diameter because it is not included in the total diameter.)


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A package states that there are 60 calories in 12 crackers and 75 calories in 15 crackers. Since the relationship is proportiona
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There are 900 calories in 180 crackers. Write 60 calories for 12 crackers as 60/12. Then make x/180, showing that there are 180 crackers with an unknown amount of calories. Cross multiply 60 and 180, which is 10,800. Divide 10,800 by 12, and you get 900.
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3 years ago
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Lines g, h, and I are parallel and
den301095 [7]

Answer: Choice C) 115 degrees

Focus on the parallel lines G and L, which are the top and bottom most horizontal lines. Notice that the transversal cut forms the alternate exterior angles 2 and 11. The angles are on the outside or exterior of the "train tracks" of the parallel lines. Also, they are on opposite sides of the transversal line. So that is how we can consider them to be alternate exterior angles.

Similar to alternate interior angles, alternate exterior angles are congruent if we have a transversal cutting a pair of parallel lines.

So in short, angle 2 and angle 11 are congruent. We are given angle 2 to be 115 degrees. Angle 11 is also 115 degrees.

3 0
3 years ago
Lify 8<br> 7x2-x+8 – 5x2 + 3x – 10
zalisa [80]

Answer: (-5x2 - 8x + 7) - (5x2 - 4)

Step-by-step explanation:

= -5x2 - 8x + 7 - 5x2 + 4

= -5x2 - 5x2 - 8x + 7 + 4

= -10x2 – 8x + 11

Answer: = (-5x2 - 8x + 7) - (5x2 - 4)

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3 years ago
Find the slope of the line through each pair of points
NNADVOKAT [17]

The formula of a slope:

m=\dfrac{y_2-y_1}{x_2-x_1}

95)

(-13,\ -10)\to x_1=-13,\ y_1=-10\\(16,\ 16)\to x_2=16,\ y_2=16

substitute:

m=\dfrac{16-(-10)}{16-(-13)}=\dfrac{16+10}{16+13}=\dfrac{26}{29}

96)

(6,\ -16)\to x_1=6,\ y_1=-16\\(-15,\ 9)\to x_2=-15,\ y_2=9

substitute:

m=\dfrac{9-(-16)}{-15-6}=\dfrac{25}{-21}=-\dfrac{25}{21}

97)

(12,\ -11)\to x_1=12,\ y_1=-11\\(14,\ -12)\to x_2=14,\ y_2=-12

substitute:

m=\dfrac{-12-(-11)}{14-12}=\dfrac{-12+11}{2}=-\dfrac{1}{2}

OK. I think you already know how to solve it. I will give short solutions now.

98)

m=\dfrac{15-(-3)}{8-(-14)}=\dfrac{18}{22}=\dfrac{9}{11}

99)

m=\dfrac{3-(-5)}{-7-3}=\dfrac{8}{-10}=-\dfrac{4}{5}

100)

m=\dfrac{-8-8}{-11-0}=\dfrac{-16}{-11}=\dfrac{16}{11}

101)

m=\dfrac{13-(-5)}{-13-(-13)}=\dfrac{18}{0}!!!-\boxed{NOT\ EXIST}

102)

m=\dfrac{18-1}{-10-2}=\dfrac{17}{-12}=-\dfrac{17}{12}

103)

m=\dfrac{19-14}{7-(-17)}=\dfrac{5}{24}

104)

m=\dfrac{12-0}{-6-4}=\dfrac{12}{-10}=-\dfrac{6}{5}

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<em><u>Look</u></em><em><u> </u></em><em><u>at</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>attached</u></em><em><u> </u></em><em><u>picture</u></em><em><u>⤴</u></em>

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