Ask question

Ask question

All categories

3 years ago

6

6(x-6)+4=8(x+2)-4 The solution is

2 answers:

vampirchik [111]3 years ago

7 0

$6(x-6)+4=8(x+2)-4\\
6x-36+4=8x+16-4\\
2x=-44\\
x=-22$

UkoKoshka [18]3 years ago

6 0

6(x-6)+4=8(x+2)-4

6(x-6)=8(x+2)-4-4

6(x-6)=8(x+2)-8

6x-36=8[(x+2)-1]

6x-36=8[x+2-1]

6x-36=8(x+1)

6x-36=8x+8

8x-6x=-36-8

2x=-44

Therefore, x=-22, as 2(-22)=-44.

You might be interested in

1.What is the simplified form of the following expression?

saul85 [17]

Answer:

WWW.MATHPAPA.COM

Step-by-step explanation:

7 0

3 years ago

PLS HELP ME W/ #8 ASAP

tiny-mole [99]

Answer:

x=2

TU:4(2)-1=7

UB: 3

TB: 10

Step-by-step explanation:

4x-1+2x-1=5x

6x-2=5x

6x=5x+2

subtract 5x from both sides

x=2

TU: (4x-1)

4(2)-1

8-1

7

TU:4(2)-1=7

UB: (2x-1)

2(2)-1

4-1

3

UB: 3

TB: 5(2)

TB: 10

7 0

3 years ago

The formula for the surface area of a right prism is LaTeX: A=2B+PhA = 2 B + P h. Solve for LaTeX: hh. Which equation gives a co

qaws [65]

Answer:

$h = \frac{A - 2B}{P}$

Step-by-step explanation:

$A = 2B + Ph$

Required

Solve for h

$A = 2B + ph$

Subtract 2B from both sides

$A - 2B= 2B - 2B+ ph$

$A - 2B= ph$

Reorder

$Ph=A - 2B$

Divide through by P

$h = \frac{A - 2B}{P}$

5 0

3 years ago

The diagonals of quadrilateral ABCD intersect at E(−2,4). ABCD has vertices at A(1,7) B(−3,5). What must be the coordinates of C

LUCKY_DIMON [66]

Answer:

The coordinates of C and D are (1, 3) and (-5, 1), respectivelly.

Step-by-step explanation:

Since E is the midpoint of diagonals AD and BC (see attachment). That is:

$AD = 2 \cdot AE$

$BC = 2\cdot BE$

The vectorial distances of AE and BE are, respectively:

$\overrightarrow{AE} = \vec E - \vec A$

$\overrightarrow {AE} = (-2,4) -(1,7)$

$\overrightarrow{AE} = (-2-1, 4-7)$

$\overrightarrow {AE} = (-3,-3)$

$\overrightarrow{BE} = \vec E - \vec B$

$\overrightarrow {BE} = (-2,4) - (-3,5)$

$\overrightarrow {BE} = (-2+3, 4-5)$

$\overrightarrow {BE} = (1,-1)$

Now, the relative vectorial distances to C and D are now obtained:

$\overrightarrow {AD} = 2\cdot \overrightarrow {AE}$

$\overrightarrow {AD} = 2 \cdot (-3,-3)$

$\overrightarrow{AD} = (-6, -6)$

$\overrightarrow {BC} = 2 \cdot \overrightarrow {BE}$

$\overrightarrow{BC} = 2 \cdot (1,-1)$

$\overrightarrow {BC} = (2,-2)$

Lastly, the coordinates are found by the following vectorial equations:

$\vec C = \vec B + \overrightarrow {BC}$

$\vec C = (-3,5) + (2,-2)$

$\vec C = (-3+2, 5 -2)$

$\vec C = (-1,3)$

$\vec D = \vec A + \overrightarrow {AD}$

$\vec D = (1,7) + (-6,-6)$

$\vec D = (1-6, 7 -6)$

$\vec D = (-5, 1)$

The coordinates of C and D are (1, 3) and (-5, 1), respectivelly.

6 0

4 years ago

Please help me on this one!

MatroZZZ [7]

On which one? You did not provide anything for me to help you with.

4 0

4 years ago