All categories

3 years ago

Drag the simplified value into the box to match each expression.

Mathematics

2 answers:

JulijaS [17]3 years ago

7 0

Can you tell me what the expressions, values and all that are
I ca't help with out the details.

Bad White [126]3 years ago

6 0

Answer:

Need more info

Step-by-step explanation:

post a pic or something.

You might be interested in

Someone please help me

timama [110]

A CANNOT DETERMINE, since you have no other data at all

6 0

3 years ago

Does the line passing through the points (-3, -6) and (1, 2) intersect the line passing through the points (-4, -1) and (-2, 3)?

Mariana [72]

Answer:

no, because when we draw then in grpah the two lines don't intersect.

8 0

2 years ago

Read 2 more answers

If f ( x ) = 2 x - 5, then f (4) is _____.<br><br> 3<br> 5<br> 4<br> 17

ICE Princess25 [194]

The answer is 3

if you substitute x as 4 then it would be 2 x 4 = 8

8-5=3

7 0

3 years ago

Read 2 more answers

Ill give you Brainlyest! I need help ASAP work shown, please!!<br> problems 8-15

8_murik_8 [283]

Using the segment addition theorem, the solutions are:

5. x = 17

6. x = 7

7. x = 14

BC = 27

CD = 61

BD = 88

8. AB = 26

9. LJ = 46

10. x = 3

11. FG = 15

12. QS = 34

13. BC = 26

14. EG = 19

15. QS = 68

<h3>How to Apply the Segment Addition Theorem?</h3>

The segment addition theorem would be used to solve the problems as shown below:

5. UW = UV + VW [segment addition theorem]

Substitute

6x - 35 = 19 + 4x - 20

6x - 4x = 35 + 19 - 20

2x = 34

x = 17

6. HJ = HI + IJ [segment addition theorem]

Substitute

7x - 27 = 3x - 5 + x - 1

7x - 3x - x = 27 - 5 - 1

3x = 21

x = 7

7. BD = BC + CD [segment addition theorem]

Substitute

7x - 10 = 4x - 29 + 5x - 9

7x - 4x - 5x = 10 - 29 - 9

-2x = -28

x = 14

BC = 4x - 29 = 4(14) - 29 = 27

CD = 5x - 9 = 5(14) - 9 = 61

BD = 7x - 10 = 7(14) - 10 = 88

8. BC = BD

Substitute

2x + 1 = 5x - 26

2x - 5x = -1 - 26

-3x = -27

x = 9

AB = 43 - BC

AB = 43 - 2x + 1 = 43 - 2(9) + 1 = 26

9. 7x - 10 = 9x - 11 - (x + 3)

7x - 10 = 9x - 11 - x - 3

7x - 9x + x = 10 - 11 - 3

-x = -4

x = 4

LJ = 28 + 7x - 10 = 28 + 7(4) - 10

LJ = 46

10. 8x + 11 = 12x - 1

8x - 12x = -11 - 1

-4x = -12

x = 3

11. 11x - 7 = 3x + 9

11x - 3x = 7 + 9

8x = 16

x = 2

FG = 11x - 7 = 11(2) - 7

FG = 15

12. 5x - 3 = 21 - x

5x + x = 21 + 3

6x = 24

x = 4

QS = 2(5x - 3) = 10x - 6 = 10(4) - 6

QS = 34

13. 8x - 20 = 2(3x - 1)

8x - 20 = 6x - 2

8x - 6x = 20 - 2

2x = 18

x = 9

BC = AB = 3x - 1 = 3(9) - 1

BC = 26

14. 5x - 1 = 7x - 13

5x - 7x = 1 - 13

-2x = -12

x = 6

EG = 6x - 4 - 13 = 6(6) - 4 - 13

EG = 19

15. RT - ST = RS

Substitute

8x - 43 - (4x - 1) = 2x - 4

8x - 43 - 4x + 1 = 2x - 4

4x - 42 = 2x - 4

4x - 2x = 42 - 4

2x = 38

x = 19

QS = 2(RS)

QS = 2(2x - 4) = 4x - 8 = 4(19) - 8

QS = 68

Learn more about the segment addition theorem on:

brainly.com/question/28263183

#SPJ1

8 0

2 years ago

21. In 4 + In(4x - 15) = In(5x + 19)

Alexxx [7]

Answer:

$x=-30;\quad \:I\ne \:0$

Step-by-step explanation:

$In\cdot \:4+In\left(4x-15\right)=In\left(5x+19\right)$

$\:4+In\left(4x-15\right):\quad -11nI+4nxI\\In\cdot \:4+In\left(4x-15\right)\\=4nI+nI\left(4x-15\right)\\\\\:In\left(4x-15\right):\quad 4nxI-15nI\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac\\a=In,\:b=4x,\:c=15\\=In\cdot \:4x-In\cdot \:15\\=4nxI-15nI\\=In\cdot \:4+4nxI-15nI\\\mathrm{Simplify}\:In\cdot \:4+4nxI-15nI:\quad -11nI+4nxI\\In\cdot \:4+4nxI-15nI\\\mathrm{Group\:like\:terms}\\=4nI-15nI+4nxI\\\mathrm{Add\:similar\:elements:}\:4nI-15nI=-11nI\\=-11nI+4nxI\\$

$\mathrm{Expand\:}In\left(5x+19\right):\quad 5nxI+19nI\\In\left(5x+19\right)\\=nI\left(5x+19\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=In,\:b=5x,\:c=19\\=In\cdot \:5x+In\cdot \:19\\=5nxI+19nI\\\\-11nI+4nxI=5nxI+19nI\\\\\mathrm{Add\:}11nI\mathrm{\:to\:both\:sides}\\-11nI+4nxI+11nI=5nxI+19nI+11nI\\Simplify\\4nxI=5nxI+30nI\\\mathrm{Subtract\:}5nxI\mathrm{\:from\:both\:sides}\\4nxI-5nxI=5nxI+30nI-5nxI\\\mathrm{Simplify}\\-nxI=30nI\\$

$\mathrm{Divide\:both\:sides\:by\:}-nI;\quad \:I\ne \:0\\\frac{-nxI}{-nI}=\frac{30nI}{-nI};\quad \:I\ne \:0\\\mathrm{Simplify}\\\frac{-nxI}{-nI}=\frac{30nI}{-nI}\\\mathrm{Simplify\:}\frac{-nxI}{-nI}:\quad x\\\frac{-nxI}{-nI}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}\\=\frac{nxI}{nI}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{xI}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=x\\\mathrm{Simplify\:}\frac{30nI}{-nI}:\quad -30\\\mathrm{Apply\:the\:fraction\:rule}:$

$\quad \frac{a}{-b}=-\frac{a}{b}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{30I}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=-30\\x=-30;\quad \:I\ne \:0$

3 0

3 years ago