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tiny-mole [99]
3 years ago
12

Solve the equation for x 2x+34=4(x+5)​

Mathematics
1 answer:
pochemuha3 years ago
8 0

Answer:

7

Step-by-step explanation:

2x+34=4(x+5)

2x+34=4x+20

-2x=-14

x=7

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he triangle on the grid will be translated two units down. On a coordinate plane, triangle A B C has points (2, 1), (0, negative
Sladkaya [172]

If these coordinates are translated two units down, we will subtract 2 units from the y axis to have A(2, -1), B(0, -3) and C(2, -3)

<h3>Translation of coordinate point</h3>

Translation is a transformation technique of changing the position of an object on the xy-plane.

Given the following coordinate of a triangle expressed as:

A(2, 1)

B(0, -1)

C(2, -1)

If these coordinates are translated two units down, we will subtract 2 units from the y axis to have:

A(2, -1)

B(0, -3)

C(2, -3)

Learn more on translation here; brainly.com/question/1046778

#SPJ1

7 0
2 years ago
Wht is the difference between 129 and 302
IceJOKER [234]
The difference is
 
 302
-129
_____
 173
to prove this
  173
+129
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3+ 9 is 12 so the unity is 2 and you plus 1 to the tenth; 7 + 2 is 9 but if you plus the number that remained in the unity is 10 so the the tenth is 0 and you will plus 1 to the hungred, 1+1 is 2 but if you plus the 1 that remained in the tenth is 3, so if we put them all together it will be 302
i hope i helped you
5 0
3 years ago
Read 2 more answers
I AM GIVING BRAINLIEST TO THE BEST ANSWER AND THE CORRECT ONE!!!!!
stiks02 [169]
The correct answer is ''b'' because 3/4 is 6/8 and 7/8 is 7/8  so also is 6/7 i hope it helped u
5 0
3 years ago
Read 2 more answers
I need the answer to this question please<br> yes i am giving out brainliest
anygoal [31]

A'(-8,-14)

B'(-5,-17)

C'(-7,-5)

4 0
3 years ago
Let $abcdefgh$ be a cube of side length 5, as shown. let $p$ and $q$ be points on $\overline{ab}$ and $\overline{ae}$, respectiv
Anna [14]
Let's buils the intersection plane:
Point P is on AB and AP=2, then PB=3; point Q is on AE and AQ=1, then QE=4. Let P' be a point on CD such that CP'=2 and Q' be a point on the plane CDHG such that P'Q'=1 and P'Q' is perpendicular to CD. The line CQ' intersects HD at point R and the plane CPQR is intersection plane.
Consider triangles ΔCDR and ΔCP'Q', they are similar. So,
\frac{CP'}{CD}=  \frac{P'Q'}{RD}  \\  \frac{2}{5}  =\frac{1}{RD}  \\ RD=2.5,
so R is a midlepoint of the side HD (for details see picture).



5 0
4 years ago
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