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3 years ago

9

Please help find volume check image also

2 answers:

Dahasolnce [82]3 years ago

6 0

Answer:

uhhhhhhhhh

Step-by-step explanation:that will bee

Rufina [12.5K]3 years ago

6 0

The formula for a triangular pyramid is $\frac{1}{3} hx^2$ .

So we have $\frac{1}{3} *9*13^{2}\\=3*169\\=507$

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Log(x^2 +5x+16) = 1 solve for x

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3 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

3 years ago