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Lyrx [107]
3 years ago
12

List the steps that you could use to solve? x 4 — = — 3 9

Mathematics
1 answer:
Lena [83]3 years ago
4 0

Answer:

  • multiply by 3

Step-by-step explanation:

\dfrac{x}{3}=\dfrac{4}{9} \qquad\text{has x-coefficient $\frac{1}{3}$}

Multiply by the reciprocal of the x-coefficient. Then you have ...

x=\dfrac{4}{3}

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One package of socks cost $7. How many packages can you but with $56?
Akimi4 [234]

Step-by-step explanation:

$7 = 1 package socks

%56 = ??

$56/$7 × 1 package socks

= 8 × 1

= 8 package socks

Hope it helps!

4 0
3 years ago
Complete the square than convert to vertex form. y=2x^2+8x-9​
8_murik_8 [283]

y=2x^2+8x-9\\D=b^2-4ac\\D=64-4(2)(-9)\\D=64+72 > 0

There are 2 roots so the only way to complete the square is,

y=2x^2+8x-9\\y=2[(x^2+4x)]-9\\y=2[(x^2+4x+4)-4]-9\\y=2[(x+2)^2-4]-9\\y=2(x+2)^2-8-9\\y=2(x+2)^2-17

Just factor 2 out of 2x^2+8x (just ignore the -9) then find the number that will make the terms be able to complete the square.

then complete the square and multiply 2 inside the brackets.

subtraction as you already get the vertex form and know how to complete the square.

Vertex Form: y=2(x+2)^2-17

3 0
3 years ago
4x-y=20 and -2x-2y=10
lara31 [8.8K]

You can use Substitution to solve this problem:


4x-y=20

-y=-4x+20

y=4x-20


Now you have two equations, and you use the first one to substitute into the second one.


y=4x-20     and     -2x-2y=10


-2x-2(4x-20)=10

-2x-8x+40=10

-10x+40=10

-10x=-30

-x=-3

x=3


Now that we have figured out what x is, we can substitute x in to one of the equations to figure out y.


4x-y=20       x=3

4(3)-y=20

12-y=20

-y=8

y=-8



<em><u>So your answer would be x=3 and y=-8</u></em>

6 0
3 years ago
A rectangular bin has a perimeter of 36 inches. The length of the bin is twice the width. Write and solve an equation to determi
bekas [8.4K]
2x + 4x = y
x=6
2(6)+4(6) = 36 sq in.
3 0
3 years ago
Can someone help me on this pls? It’s urgent, so ASAP (it’s geometry)
GarryVolchara [31]

<u>Question 6</u>

1) \overline{AB} \cong \overline{BD}, \overline{CD} \perp \overline{BD}, O is the midpoint of \overline{BD}, \overline{AB} \cong \overline{CD} (given)

2) \angle ABO, \angle ODC are right angles (perpendicular lines form right angles)

3) \triangle ABO, \triangle CDO are right triangles (a triangle with a right angle is a right triangle)

4) \overline{BO} \cong \overline{OD} (a midpoint splits a segment into two congruent parts)

5) \triangle ABO \cong \triangle CDO (LL)

<u>Question 7</u>

1) \angle ADC, \angle BDC are right angles), \overline{AD} \cong \overline{BD}

2) \overline{CD} \cong \overline{CD} (reflexive property)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

4) \triangle ADC \cong \triangle BDC (LL)

5) \overline{AC} \cong \overline{BC} (CPCTC)

<u>Question 8</u>

1) \overline{CD} \perp \overline{AB}, point D bisects \overline{AB} (given)

2) \angle CDA, \angle CDB are right angles (perpendicular lines form right angles)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

4) \overline{AD} \cong \overline{DB} (definition of a bisector)

5) \overline{CD} \cong \overline{CD} (reflexive property)

6)  \triangle ADC \cong \triangle BDC (LL)

7) \angle ACD \cong \angle BCD (CPCTC)

8 0
2 years ago
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