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3 years ago

15

Which two-dimensional figure could be a cross section of a rectangular pyramid that has been intersected by a plane perpendicula

r to its base and through its vertex?

1 answer:

DanielleElmas [232]3 years ago

7 0

The choices are found elsewhere and the figures would be:
a. rectangleb. trianglec. squared. trapezoid
From the choices, the answer would be option B. A triangle figure would be the cross section when <span>a rectangular pyramid that has been intersected by a plane perpendicular to its base and through</span>its vertex.

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In the diagram which angles pat of a linear sair and part

Bas_tet [7]

Answer:

The word linear goes back into mathematics and algebra. . I have taken multiple courses over the years. You don't have to tell me the course your in. But are you trying to use mathematics geometry or algebra

Step-by-step explanation:

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3 years ago

BC has endpoints B(-3,-5) and C(12,12). FInd the coordinates of the midpoint of BC

forsale [732]

Answer:

(9/2, 7/2)

Step-by-step explanation:

The easiest way to remember how to do this problem is to average the x-coordinates and average the y-coordinates.

So, in your problem (-3 + 12)/2 = 9/2 and (-5 + 12)/2 = 7/2

The midpoint is (9/2, 7/2)

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2 years ago

I know how to do it I'm just lazy

alexandr402 [8]

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3 years ago

The sum of a number and 9 is multiplied by -2 and the answer is -8

nadya68 [22]

9(-2)+x=-8
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3 years ago

For 0 ≤ ϴ < 2π, how many solutions are there to tan(StartFraction theta Over 2 EndFraction) = sin(ϴ)? Note: Do not include va

Black_prince [1.1K]

Answer:

3 solutions:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

Step-by-step explanation:

So, first of all, we need to figure the angles that cannot be included in our answers out. The only function in the equation that isn't defined for some angles is $tan(\frac{\theta}{2})$ so let's focus on that part of the equation first.

We know that:

$tan(\frac{\theta}{2})=\frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})}$

therefore:

$cos(\frac{\theta}{2})\neq0$

so we need to find the angles that will make the cos function equal to zero. So we get:

$cos(\frac{\theta}{2})=0$

$\frac{\theta}{2}=cos^{-1}(0)$

$\frac{\theta}{2}=\frac{\pi}{2}+\pi n$

or

$\theta=\pi+2\pi n$

we can now start plugging values in for n:

$\theta=\pi+2\pi (0)=\pi$

if we plugged any value greater than 0, we would end up with an angle that is greater than $2\pi$ so, that's the only angle we cannot include in our answer set, so:

$\theta\neq \pi$

having said this, we can now start solving the equation:

$tan(\frac{\theta}{2})=sin(\theta)$

we can start solving this equation by using the half angle formula, such a formula tells us the following:

$tan(\frac{\theta}{2})=\frac{1-cos(\theta)}{sin(\theta)}$

so we can substitute it into our equation:

$\frac{1-cos(\theta)}{sin(\theta)}=sin(\theta)$

we can now multiply both sides of the equation by $sin(\theta)$

so we get:

$1-cos(\theta)=sin^{2}(\theta)$

we can use the pythagorean identity to rewrite $sin^{2}(\theta)$ in terms of cos:

$sin^{2}(\theta)=1-cos^{2}(\theta)$

so we get:

$1-cos(\theta)=1-cos^{2}(\theta)$

we can subtract a 1 from both sides of the equation so we end up with:

$-cos(\theta)=-cos^{2}(\theta)$

and we can now add $cos^{2}(\theta)$

to both sides of the equation so we get:

$cos^{2}(\theta)-cos(\theta)=0$

and we can solve this equation by factoring. We can factor $cos(\theta)$ to get:

$cos(\theta)(cos(\theta)-1)=0$

and we can use the zero product property to solve this, so we get two equations:

Equation 1:

$cos(\theta)=0$

$\theta=cos^{-1}(0)$

$\theta={\frac{\pi}{2}, \frac{3\pi}{2}}$

Equation 2:

$cos(\theta)-1=0$

we add a 1 to both sides of the equation so we get:

$cos(\theta)=1$

$\theta=cos^{-1}(1)$

$\theta=0$

so we end up with three answers to this equation:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

7 0

2 years ago