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3 years ago

13

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

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