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2 years ago

Solve the compound inequality: 3≤6−3x<15

Mathematics

1 answer:

DerKrebs [107]2 years ago

4 0

solve 3≤−3x+6<153≤−3x+6<15

Answer: (−3,1](−3,1]

Approximate Form: (−3,1]

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Which statements are true? Select all that apply.

xeze [42]

Answer:

A and C

Step-by-step explanation:

4 0

2 years ago

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The product of two consecutive odd integers is 1 less than four times their some. find the two integers.

nignag [31]

2 consecutive odd integers.....x and x + 2.....the product means multiply....the sum means add.

x(x + 2) = 4(x + x + 2) - 1
x^2 + 2x = 4(2x + 2) - 1
x^2 + 2x = 8x + 8 - 1
x^2 + 2x = 8x + 7
x^2 + 2x - 8x - 7 = 0
x^2 - 6x - 7 = 0
(x - 7)(x + 1) = 0

x - 7 = 0 x + 2 = 7 + 2 = 9
x = 7

x + 1 = 0 x + 2 = -1 + 2 = 1
x = -1

there can be 2 answers for this....
(1) 7 and 9
(2) -1 and 1

7 0

2 years ago

The sum of a number and 6 is 8 more than twice the number. Find the equation that could be used to find this number .

Reil [10]

The correct answer is D

6 0

2 years ago

2/3 + 3/4=? (pls show the steps on how you got an awnser :) )

shepuryov [24]

Answer:

17/12= 1 5/12

Step-by-step explanation:

Common denominator:

2/3= 8/12

3/4= 9/12

Solve:

8+9= 17

17/12= 1 5/12

8 0

2 years ago

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Cos^2x+cos^2(120°+x)+cos^2(120°-x) i need this asap. pls help me

o-na [289]

Answer:

$\frac{3}{2}$

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

---------------------------------------------------------------

cos(120 + x) = cos120cosx - sin120sinx

= - cos60cosx - sin60sinx

= - $\frac{1}{2}$ cosx - $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos² (120 + x)

= $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------

cos(120 - x) = cos120cosx + sin120sinx

= -cos60cosx + sin60sinx

= - $\frac{1}{2}$ cosx + $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos²(120 - x)

= $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------------

Putting it all together

cos²x + $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x + $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

= cos²x + $\frac{1}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ (cos²x + sin²x) = $\frac{3}{2}$

5 0

2 years ago