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Soloha48 [4]
3 years ago
13

How many lines of symmetry does the shape have?

Mathematics
2 answers:
givi [52]3 years ago
8 0

Answer:

the answer is two

Step-by-step explanation:

top left to bottom right and top right to bottom left

:)

nlexa [21]3 years ago
3 0
I’m pretty sure 4. I could be wrong but it has 4 angles, so it could be.
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Find the amount (future value) of the ordinary annuity. (round your answer to the nearest cent.) $1500/semiannual period for 8 y
mixer [17]
The formula of the future value of an annuity ordinary is
Fv=pmt [((1+r/k)^(kn)-1)÷(r/k)]
Fv future value?
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R interest rate 0.025
K compounded semiannual 2
N time 8 years
Fv=1,500×(((1+0.025÷2)^(2×8)
−1)÷(0.025÷2))
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Hope it helps!
6 0
3 years ago
Please help and explain why!
scoundrel [369]
1) The measures of the opposite angles of an inscribed quadrilateral are supplementary, add up to 180.

That means that angle ROP and angle RQP are supplementary.

We can add them and set them equal to 180 so that we can solve for x first.

∠ROP + RQP = 180
x + 17 + 6x - 5 = 180
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7x = 168
x = 24

Now use the 24 for x and solve for the measure of angle ROP.

∠ROP = x + 17
∠ROP = 24 + 17
∠ROP = 41

The answer is A

2) The answer is A.

Hope this helps :)
4 0
3 years ago
Solve by using the perfect squares method.
Julli [10]
X² + 16x + 64 = 0
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x + 8 = √0

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3 0
3 years ago
Read 2 more answers
Write a recursive definition for the sequence 14, 10, 6, 2...
VladimirAG [237]
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a_{n}=a_{1}+d(n-1)

So, what we need to do is to find a_{1} and d.
a_{1}=14
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Having found these two values we can now define our recursive sequence:
a_{n}=14-4(n-1)
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7 0
3 years ago
Read 2 more answers
Jamie is riding a Ferris wheel that takes fifteen seconds for each complete revolution. The diameter of the wheel is 10 meters a
Agata [3.3K]

Answer:

The answers to the question is

(a) Jamie is gaining altitude at 1.676 m/s

(b) Jamie rising most rapidly at t = 15 s

At a rate of 2.094 m/s.

Step-by-step explanation:

(a) The time to make one complete revolution = period T = 15 seconds

Here will be required to develop the periodic motion equation thus

One complete revolution = 2π,

therefore the  we have T = 2π/k = 15

Therefore k = 2π/15

The diameter = radius of the wheel = (diameter of wheel)/2 = 5

also we note that the center of the wheel is 6 m above ground

We write our equation in the form

y = 5*sin(\frac{2*\pi*t}{15} )+6

When Jamie is 9 meters above the ground and rising we have

9 = 5*sin(\frac{2*\pi*t}{15} )+6 or 3/5 = sin(\frac{2*\pi*t}{15} ) = 0.6

which gives sin⁻¹(0.6) = 0.643 =\frac{2*\pi*t}{15}

from where t = 1.536 s

Therefore Jamie is gaining altitude at

\frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi t}{15}) = 1.676 m/s.

(b) Jamie is rising most rapidly when   the velocity curve is at the highest point, that is where the slope is zero

Therefore we differentiate the equation for the velocity again to get

\frac{d^2y}{dx^2} = -5*(\frac{\pi *2}{15} )^2*sin(\frac{2\pi t}{15}) =0, π, 2π

Therefore -sin(\frac{2\pi t}{15} ) = 0 whereby t = 0 or

\frac{2\pi t}{15} = π and t =  7.5 s, at 2·π t = 15 s

Plugging the value of t into the velocity equation we have

\frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi t}{15}) = - 2/3π m/s which is decreasing

so we try at t = 15 s and we have \frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi *15}{15}) = \frac{2}{3} \pim/s

Hence Jamie is rising most rapidly at t = 15 s

The maximum rate of Jamie's rise is 2/3π m/s or 2.094 m/s.

7 0
3 years ago
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