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Travka [436]
4 years ago
5

Find x. please help!

Mathematics
1 answer:
elena55 [62]4 years ago
8 0
Both sides left and right are equal because they have equal angles.

2x-5=x+40

x=45.

plug in to be sure.

2(45)-5= 85
&
(45)+40=85
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Joe is putting up a rectangular fence in his back yard to contain his dog, murphy . If joe bought 98 feet of fencing and one dim
NeX [460]

Answer:

29 ft.

Step-by-step explanation:

Since one dimension of the yard is 20 ft., we multiply that by 2 (20 × 2 = 40) to get both sides of the yard with that dimension. Then, we subtract 98 - 40 = 58. 58 ft. is the total of BOTH of the other two sides, so we divide that by 2 (58 ÷ 2 = 29) to get the final answer.

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3 years ago
How to find the median of <br><br> E)6,10,12,5,7,12,9
emmasim [6.3K]
First you order them from least to greatest.

5, 6, 7, 9, 10, 12, 12

Then you cross off one number from each side until you get to the middle

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6 0
4 years ago
Read 2 more answers
Hey guys I have a question!
Alenkasestr [34]
It kinda depends on the grade scaling and what percent of your grade test are worth. But I'm guessing it would drop down to a 70% or more.
4 0
3 years ago
In one week 97 cell phones were sold. The following week, 48 more cell phones were sold than the week before. How many cell phon
baherus [9]

145. 7 + 8 is 15 carry the 1, 9+4+1 is 14, so 145

8 0
3 years ago
Find an equation of the plane. The plane that passes through the line of intersection of the planes x − z = 3 and y + 2z = 3 and
IgorLugansk [536]

Answer:

x-y-3=0

Step-by-step explanation:

Choose two common points on the line of intersection of the planes x − z = 3 and y + 2z = 3:

  • 1st point: if x=0, then z=-3 and y=3-2z=3+6=9, so the 1st point is (0,-3,9);
  • 2nd point: if x=3, then z=0 and y=3-2z=3-0=3, so the 2nd point is (3,0,3).

The perpendicular plane to the plane x+y-4z=6 is parallel to the vector with coordinates (1,1,-4) (normal vector of the given plane).

Hence, the equation of needed plane is

\left\|\begin{array}{ccc}x-0&y-(-3)&z-9\\3-0&0-(-3)&3-9\\1&1&-4\end{array}\right\|=0\Rightarrow \left\|\begin{array}{ccc}x&y+3&z-9\\3&3&-6\\1&1&-4\end{array}\right\|=0

Thus,

\left\|\begin{array}{ccc}x&y+3&z-9\\3&3&-6\\1&1&-4\end{array}\right\|=0\Rightarrow \\\\-12x-6(y+3)+3(z-9)-3(z-9)+6x+12(y+3)=0\\ \\-6x+6(y+3)=0\\ \\-x+y+3=0\\ \\x-y-3=0

4 0
4 years ago
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