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a_sh-v [17]
3 years ago
8

Nathaniel builds birds and birdhouses using Lego blocks. Let BBB represent the number of birds and HHH represent the number of b

irdhouses that Nathaniel can build with his Lego blocks. 43B+215H \leq 300043B+215H≤300043, B, plus, 215, H, is less than or equal to, 3000 Nathaniel wants to build 505050 birds using Lego blocks. How many birdhouses can he build at most with the remaining Lego blocks? Choose 1 answer: Choose 1 answer: (Choice A) A Nathaniel can build at most 111 birdhouse. (Choice B) B Nathaniel can build at most 222 birdhouses. (Choice C) C Nathaniel can build at most 333 birdhouses. (Choice D) D Nathaniel can build at most 444 birdhouses.
Mathematics
1 answer:
vladimir1956 [14]3 years ago
8 0

Answer:

(C) Nathaniel can build at most 3 birdhouses.

Step-by-step explanation:

Given that:

43B+215H \leq 3000

where:

  • B represents the number of birds
  • H represents the number of Birdhouses

Nathaniel wants to build 50 birds(B) using lego Blocks, we want to determine how many birdhouses(H) he can build with the remaining Lego blocks.

If B=50

43(50)+215H \leq 3000\\2150+215H \leq 3000\\\text{Subtract 2150 from both sides}\\215H \leq 3000-2150\\215H \leq 850\\\text{Divide both sides by 215}\\H  \leq 3.95

Therefore, Nathaniel can build at most 3 Birdhouses.

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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
3 years ago
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Alja [10]

Answer:

55

Step-by-step explanation:

-5 (- 11)

= 55

5 0
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