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never [62]
3 years ago
13

What expression is equivalent to 3(x-2)+2x

Mathematics
1 answer:
WINSTONCH [101]3 years ago
4 0

Answer:5x - 6

Step-by-step explanation:

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An equilateral triangle has a perimeter of 15x + 33x feet. What is the length of each side?
mart [117]

Answer:

16

Step-by-step explanation:

15x+33x =48x

Since there are 3 equal sides, you need to divide 48x by 3.

Hope this helps! :)

6 0
4 years ago
Find the area of triangle ABC with the given parts. Round to the nearest whole number.
slavikrds [6]
\sin{26.4^o}= \frac{h}{12.3} 
\\
\\h=12.3\times \sin{26.4^o}=5.47
\\
\\A= \frac{7.7\times5.47}{2}= 21.06

3 0
4 years ago
1/4 ( x − 12 ) + 1/3 ( x + 9 )
navik [9.2K]

Answer: 7/12x

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Which expression can be used to convert 80 US dollars Australian dollars
andreev551 [17]

80 USD × (1.0343 AUD / 1 USD)

Solution:

To convert 80 US dollars into Australian dollars.

1 US dollar = 1.0343 AU dollar

1 AU dollar = 0.9668 US dollar

To find 80 US dollars into Australian dollars multiply 80 US dollars by 1.0343

80 AU dollars = 80 × 1.0343/ 1 USD

The expression is 80 USD × (1.0343 AUD / 1 USD)

Hence, the expression which is used to convert 80 US dollars to Australian dollars is 80 USD × (1.0343 AUD / 1 USD).

7 0
4 years ago
Find all the zeros of the equation. -x^3-3x^2=6x+4
k0ka [10]

Answer:

The zeros of given expression -x^3-3x^2=6x+4 is -1, -1+i\sqrt{3} and -1-i\sqrt{3}

Step-by-step explanation:

Given expresssion is -x^3-3x^2=6x+4

To find zeros of given expression we have to equate the expression to zero.

ie., -x^3-3^2-6x-4=0

-(x^3+3x^2+6x+4)=0

x^3+3x^2+6x+4=0

By using synthetic division

-1  |  1    3    6    4

    |  0   -1   -2   -4

    |________________

       1    2   4    0

Therefore (x+1) is a  zero

Now the quadratic equation is x^2+2x+4=0

For quadratic equation ax^2+bx+c=0  we have

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Here a=1 ,b=2 and c=4 now substitute the values

x=\frac{-2\pm \sqrt{2^2-4(1)(4)}}{2(1)}

x=\frac{-2\pm \sqrt{4-16}}{2}

x=\frac{-2\pm \sqrt{-12}}{2}

x=\frac{-2\pm \sqrt{12i^2}}{2} where i^2=-1

x=\frac{-2\pm i \sqrt{4\times 3}}{2}

x=\frac{-2\pm i \sqrt{4}\sqrt{3}}{2}

x=\frac{-2\pm 2i\sqrt{3}}{2}

x=2\times\frac{(-1\pm i\sqrt{3})}{2}

x=-1\pm i\sqrt{3}

Therefore x=-1+i\sqrt{3}  and x=-1-i\sqrt{3}

Therefore the zeros are -1, -1+i\sqrt{3} and -1-i\sqrt{3}

4 0
4 years ago
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