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topjm [15]
1 year ago
9

Solve inequality for x -4(2x+4)>8x+64

Mathematics
2 answers:
AlladinOne [14]1 year ago
8 0

Answer:

x < -40/13

Step-by-step explanation:

-4(2x + 4) > 8x + 64

-8x -16 > 8x + 64              use distributive property first

+8x         +8x                    isolate the variable to one side

-16 > 26x + 64

-64           -64                 -64 from both sides to get both terms on either side

-80 > 26x

÷26   ÷26                       ÷26 on both sides to get x without any numbers

-80/26 > x

x < -80/26                    

x < -40/13                       simplify

Your answer is x < -40/13

make sure the negative symbol is not next to the 40 only but is where the line is to show the fraction (I hope this makes sense - let me know if not)

Hope this helps!

                                             

Mkey [24]1 year ago
4 0

Answer:

x< - \frac{16}{3}

Step-by-step explanation:

x 8x_16>8x+64

Add similar elemants:x - 8x= -7x

-7x - 16 > 8x + 64

Add 16 to both sides

-7x- 16 + 16 > 8x + 64 + 16

Simplify

- 7x > 8x + 80

Subtract 8x from both sides

-7x - 8x > 8x + 80 - 8x

Simplify

-15x > 80

Multiply both sides by - 1 ( reverse the inequality)

(-15x) ( - 1) < 80 ( - 1 )

simplify

\frac{15x}{15} < \frac{-80}{15}

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Allushta [10]

Answer:

Step-by-step explanation:

9) PQR Is an isosceles triangle

=> ∠PRQ = (180° - x)/2

PRS is an isosceles right triangle

=> ∠PRS = 45°

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=> \frac{180-x}{2}+45=115 \\

=> 180 - x = (115 - 45).2 = 140

<=> x = 180 - 140 = 40

10) ABD is an isosceles right triangle => ∠ABD = 45°

BCD is an equilateral triangle => ∠CBD = 60°

have: x = ∠ABD + ∠CBD = 45° + 60° = 105°

11) have: x = y (2)

PQT is an isosceles triangle => ∠PQT = 180 - 70.2 = 40

QTS   is an isosceles triangle => ∠TQS = 180 -2x

QRS   is an isosceles triangle => ∠RSQ = y

have: 40 + 180 - 2x + y = 180 => 2x - y = 40 (1)

(1)(2) => \left \{ {{y=x} \atop {2x-y=40}} \right.\\\\=> \left \{ {{x=40} \atop {y=40}} \right.

=> x + y = 80

12) EFJ Is an equilateral  triangle => ∠FJE = 60

∠FJE is the outer angle of the triangle FHJ but FHJ is an isosceles triangle

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∠JHF is the outer angle of the triangle FHG

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3 years ago
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Answer:

  • volume: 252 cm³
  • mass: 2167.2 (units not specified)

Step-by-step explanation:

The volume of a cuboid is given by the product of its dimensions. Here, that is ...

  (12 cm)·(6 cm)·(3.5 cm) = 252 cm³

We are told the mass of each cm³ is 8.6, so 252 of them will have a mass of ...

  8.6·252 cm³ = 2167.2 . . . . . no units specified

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3 years ago
It is given that the straight line y = (q + 2)x + 10 is parallel to the straight line 4x + 2y = 5. Find the value of q. ​
solmaris [256]

Answer:

q= -4

Step-by-step explanation:

First, simplify 4x+2y=5

2y=-4x+5

y=-2x+5/2

q+2 has to equal to -2

q = -4

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2 years ago
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You have a fishing line spool with an end that has an area of 20.0 cm2. how much fishing line do you need to wind around the spo
grandymaker [24]

Answer: 158.53 cm

Step-by-step explanation:

We know that the ends of a fishing line spool are circular in shape.

Given : The area of an end of fishing line spool = 20.0\cm^2   (1)

Area of a circle = \pi r^2                                (2)

Circumference of a circle = 2\pi r                           (3)

, where r is radius of the circle.

From (1) and (2), we have

\pi r^2=20\\\\\Rightarrow\ r^2=\dfrac{20}{\pi}\\\\\Rightarrow\ r=\sqrt{\dfrac{20}{\pi}}                

Circumference of fishing spool = 2\pi r               (using (3))

=2\pi \sqrt{\dfrac{20}{\pi}}=2\sqrt{20\pi}=2\times2\sqrt{5\pi}=4\sqrt{5\pi}

i.e. Fishing spool required to wind around the spool one time =4\sqrt{5\pi}\ cm

⇒ Fishing spool required to wind around the spool 10 times =10\times4\sqrt{5\pi}\ cm=40\sqrt{5\pi}\\\\=40\times\sqrt{5}\times \pi\\\\=40\times2.236\times\sqrt{3.14159}\\\\=89.44\times1.772453\\\\=158.528205473\approx158.53\ cm

Hence, you need 158.53 cm or about 159 cm of fishing line to wind around the spool 10 times.

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