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liq [111]
3 years ago
12

Todea what did one pencil say to another mathematics algebra order of operations answer

Mathematics
1 answer:
Serjik [45]3 years ago
8 0
Part 1

6- \frac{18-3^2}{4+(2-3)} =6- \frac{18-9}{4+(-1)}  \\  \\ =6- \frac{9}{4-1} =6- \frac{9}{3} =6-3 \\  \\ =3



Part 2

7-8\div4(3^2-1)=7-8\div4(9-1) \\  \\ =7-8\div4(8)=7-8\div32=7- \frac{1}{2} \\  \\ =6  \frac{1}{2} = \frac{13}{2}



Part 3

2-2(2-5)^2+3^2=2-2(-3)^2+9 \\  \\ =2-2(9)+9=2-18+9=-7



Part 4

(5^2-9\cdot3)^2-11=(25-27)^2-11 \\  \\ =(-2)^2-11=4-11=-7



Part 5

[(2-3)^2-5]\cdot4=[(-1)^2-5]\cdot4 \\  \\ =(1-5)\cdot4=-4\cdot4=-16



Part 6

12-11\cdot2+16\div8=12-22+2 \\  \\ =-8



Part 7

-5+1\cdot3-(7-2^3)=-5+3-(7-8) \\  \\ =-2-(-1)=-2+1=-1



Part 8

8+2\cdot3-14\div7=8+6-2 \\  \\ =12



Part 9

(8-2)^2+\frac{1}{4}[4-3(6-10)]=6^2+\frac{1}{4}[4-3(-4)] \\  \\ =36+\frac{1}{4}[4-(-12)]=36+\frac{1}{4}(4+12)=36+\frac{1}{4}(16) \\  \\ =36+4=40



Part 10

7+(2-3^2)\cdot5=7+(2-9)\cdot5 \\  \\ =7+(-7)\cdot5=7+(-35)=7-35 \\  \\ =-28



Part 11

3-2[2^3+(-1)^3]=3-2[8+(-1)] \\  \\ =3-2(8-1)=3-2(7)=3-14 \\  \\ -11



Part 12

18+6\div3(2^2+5)=18+6\div3(4+5) \\  \\ =18+6\div3(9)=18+6\div27=18+ \frac{2}{9} \\  \\  =18 \frac{2}{9}
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Answer:

x = 15.3°

Step-by-step explanation:

Trig identities:

sin(x)=\dfrac{O}{H} \ \ \ \ \ cos(x)=\dfrac{A}{H} \ \ \ \ \ tan(x)=\dfrac{O}{A}

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8 0
2 years ago
If E F = 7 , A C = 16 , and A F = 9
Vilka [71]

Answers:

CB = 14

GF = 8

FB = 9

EF is parallel to CB

====================================

Explanations:

Points E and F are midpoints of their respective sides. They form the midsegment EF. Because EF is a midsegment, A midsegment is half the length of its parallel counterpart, so CB is two times longer than EF. If EF is 7 units long, then CB = 2*EF = 2*7 = 14

For similar reasons, GF is parallel to AC. If AC = 16, then half of that is GF = (1/2)*AC = 0.5*16 = 8.

FB = FA = 9 as these segments have the same single tickmark to indicate they are the same length

EF is parallel to CB because EF is a midsegment, and this is one of the properties of being a midsegment. We can show that quadrilateral EGBF is a parallelogram to help prove this.

8 0
3 years ago
The line AB has midpoint (2,5).<br> A has coordinates (1, 2).<br> Find the coordinates of B.
Gekata [30.6K]

Answer:

X_m = \frac{A_x +B_x}{2}= \frac{1+B_x}{2}= 2

And we can solve for B_x and we got:

1+B_x = 4

B_x = 3

Y_m = \frac{A_y +B_y}{2}= \frac{2+B_y}{2}= 5

And we can solve for B_x and we got:

2+B_y = 10

B_y = 8

So then the coordinates for B are (3,8)

Step-by-step explanation:

For this case we know that the midpoint for the segment AB is (2,5)

And we know that the coordinates of A are (1,2)

We know that for a given segment the formulas in order to find the midpoint are given by:

X_m = \frac{A_x +B_x}{2}= \frac{1+B_x}{2}= 2

And we can solve for B_x and we got:

1+B_x = 4

B_x = 3

Y_m = \frac{A_y +B_y}{2}= \frac{2+B_y}{2}= 5

And we can solve for B_x and we got:

2+B_y = 10

B_y = 8

So then the coordinates for B are (3,8)

7 0
3 years ago
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