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klio [65]
2 years ago
5

Which describes how triangle D was transformed to result in similar triangle D'?

Mathematics
2 answers:
I am Lyosha [343]2 years ago
6 0

Answer:

C. Reflection

Step-by-step explanation:

Hope it helps!

Tpy6a [65]2 years ago
5 0
It id c good luck:) :)
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Please help. I am doing a test and I do not understand what I am being asked.
Oksanka [162]

Answer:

one

Step-by-step explanation:

i have no idea what kind of test ur taking my guy but have fun

5 0
3 years ago
My noob has 36 apples he puts 29 in a bag how much he has now
SCORPION-xisa [38]

Answer: He has 7 at the moment, technically, if he still has the bag, he still has all 36

Step-by-step explanation:

8 0
3 years ago
Morgan works at a store that sells clothing accessories. The supply function
jarptica [38.1K]

Answer:

The number of supply of base balls is 22

Step-by-step explanation:

Given

P = Q - 4

Price = $18

Required

Number of supply

The relationship between price and quantity is given to be P = Q - 4 where price is represented by P and Q represents the quantity.

To get the quantity supply when price is $18, all you need to do is to substitute 18 for P in the above equation;

Thus, giving:

18 = Q - 4

Make Q the subject of formula

Q = 18 + 4

Q = 22 quantities

Hence,  the number of supply of base balls is 22

6 0
3 years ago
Read 2 more answers
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
It took a race car driver 3 hours and 15 minutes to go 500 miles. What was his mean rate of speed, to the nearest tenth of a mil
WINSTONCH [101]
In order to calculate rate or average speed you divide miles by time

I got 153.8 by dividing 500 by 3.25 (the .25 is because 15 minutes is 1/4 of 60) hope this helped!!
4 0
3 years ago
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