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3 years ago

Which statement is true about the points and planes

Mathematics

2 answers:

s2008m [1.1K]3 years ago

8 0

<h2>Answer:</h2>

The statement that is true about the point and planes is:

The line that can be drawn through points D and E is contained in plane γ.

<h2>Step-by-step explanation:</h2>

We know that if two points lie in a plane that the line connecting those two points will also lie in the same plane and hence all the points above that line are in the same plane.

and if two points are in different plane which are perpendicular than the line connecting those two points lie outside both the planes.

Hence, from the above points we have:

The line that can be drawn through points D and E is contained in plane γ.

Anarel [89]3 years ago

5 0

both the second and fourth ones look right to me, but i'm pretty sure its the fourth option.

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Here are two rectangles.

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Answer:

$AB=9.5\ cm$

Step-by-step explanation:

step 1

Find the length side PQ

we know that

The area of rectangle PQRS is given by

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2 years ago

Use the tangent identity to

Lunna [17]

Answer:

$tan(x)=\frac{44\sqrt{89}}{89}$

Step-by-step explanation:

tan is defined as: $tan(x)=\frac{opposite}{adjacent}$

sin is defined as: $sin(x)=\frac{opposite}{hypotenuse}$

cos is defined as: $cos(x)=\frac{adjacent}{hypotenuse}$

We can also define tan as: $tan(x)=\frac{sin(x)}{cos(x)}$

because plugging in the definitions of sin and cos in we get: $tan(x)=\frac{(\frac{opposite}{hypotenuse})}{(\frac{adjacent}{hypotenuse})}\\\\tan(x)=\frac{opposite}{hypotenuse}*\frac{hypotenuse}{adjacent}\\\\tan(x)=\frac{opposite}{adjacent}$

which you'll notice is the original definition of tan(x)

So using this definition of tan(x) we can use the givens sin(x) and cos(x) to find tan(x)

$sin(x)=\frac{44}{45}\\\\cos(x)=\frac{\sqrt{89}}{45}\\\\tan(x)=\frac{sin(x)}{cos(x)}$

plugging in sin(x) and cos(x) we get:

$tan(x)=\frac{\frac{44}{45}}{\frac{\sqrt{89}}{45}}\\\\tan(x)=\frac{44}{45}*\frac{45}{\sqrt{89}}\\\\tan(x)=\frac{44}{\sqrt{89}}$

We usually don't like square roots in the denominator, and from here we want to rationalize the denominator which we do by removing the square root from the denominator.

We can do this by multiplying the fraction by: $\frac{\sqrt{89}}{\sqrt{89}}$ which doesn't change the value of the fraction since it simplifies to 1, but it gets rid of the square root in the denominator

$tan(x)=\frac{44}{\sqrt{89}}*\frac{\sqrt{89}}{\sqrt{89}}\\\\tan(x)=\frac{44\sqrt{89}}{89}$

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7 months ago