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2 years ago

13

Juan is making a model out of rhombi. Each rhombus will be connected to the one before it. Each rhombus will be 6 inches tall an

d 4 inches wide.

1 answer:

bazaltina [42]2 years ago

8 0

What is the exact question here ?

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Find all the zeros of the equation

lapo4ka [179]

Divide both sides by -3, and replace $x^2$ with $y$ . Then

$-3x^4+27x^2+1200=0\iff y^2-9y-400=0$

Factorize the quadratic in $y$ to get

$y^2-9y-400=(y+16)(y-25)=0\implies y=-16,y=25$

which in turn means

$x^2=-16,x^2=25$

But $x^2\ge0$ for all real $x$ , so we can ignore the first solution. This leaves us with

$x^2=25\implies x=\pm\sqrt{25}=\pm5$

If we allow for any complex solution, then we can continue with the solution we ignored:

$x^2=-16\implies x=\pm\sqrt{-16}=\pm i\sqrt{16}=\pm4i$

6 0

3 years ago

Which number line represents the solutions to |x - 2|= 6?

Alexeev081 [22]

Answer:8 and -4

Step-by-step explanation:

[x-2]=6 ,x-2 =6 ,x=8 and x-2=-6 ,x=-4

4 0

2 years ago

Find the value of X<br><br> please help me its due in an hour

spayn [35]

Answer:

Step-by-step explanation:

4 0

2 years ago

3u+5u=64<br> What is the simplify of this <br> What does u

Nutka1998 [239]

Answer:

u = 8

Step-by-step explanation:

3u + 5u = 64

8u = 64

8u/8 = 64/8

u = 8

8 0

3 years ago

What is the exact value of arccos(−2√2) ?

Strike441 [17]

Answer:

See explanation

Step-by-step explanation:

The range of the cosine function is $-1\le \cos x\le 1$

Therefore $arc \cos (-2\sqrt{2})$ is not defined.

Assuming your question is rather $arc \cos (-\frac{\sqrt{2}}{2})$

Then $arc \cos (-\frac{\sqrt{2}}{2})=\pi-arc \cos (\frac{\sqrt{2}}{2})$ in the second quadrant.

$arc \cos (-\frac{\sqrt{2}}{2})=\pi-\frac{\pi}{4}$

$arc \cos (-\frac{\sqrt{2}}{2})=\frac{3\pi}{4}$

4 0

2 years ago