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PolarNik [594]
3 years ago
5

Which graph shows the line y-1=2(x+2)

Mathematics
2 answers:
Yanka [14]3 years ago
3 0
Graph C. The equation of the line is y=2x+5. 5 is the y-intercept so the line has to go through that point.
garri49 [273]3 years ago
3 0

Answer:

A) Graph C

Step-by-step explanation:

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-6x+4+3x-2?
Dovator [93]

Answer:

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
#6 :) :) how do you do this ?????
Romashka-Z-Leto [24]
The common difference is -5
right now, a₁₂ = 22

to get a₁, subtract 5 each time you subtract one from the subscript

so

a₁₁ = 17
a₁₀ = 12
a₉ = 7
a₈ = 2
a₇ = -3
a₆ = -8
a₅ = -13
a₄ = -18
a₃ = -23
a₂ = - 28
a₁ = -33



a₁ = -33

hope this helps
8 0
3 years ago
What is the prime factorization of 11
daser333 [38]

Answer:

11=11*1

Step-by-step explanation:

Eleven is a prime number, you can see that by proving with other minor prime numbers. like 2,3, 5 and 7

Neighter of those divide exactly 11, except 1

3 0
4 years ago
The cylinders are similar. The volume of the larger cylinder is 9648 cubic inches. What is the volume of the smaller cylinder?
zalisa [80]
Hey there :)

The volume formula of a cylinder is:
V = \pir²h

The volume of the larger volume is given to be 9648³ and the height is 18 in
We need to find the radius first

9648 = \pir²(18)
\frac{9648}{18} = \pir²
536 = \pir²
\frac{536}{ \pi} = r²
\sqrt{ \frac{536}{ \pi } } = r
r ≈ 13.06

The ratio of the big cylinder to small cylinder is:
18 : 9   (given height)
So the big cylinder is 2 times bigger than the small cylinder
Therefore the radius of the big cylinder is as well 2 times bigger

\frac{13.06}{2} = r
r ≈ 6.53  ( the radius of the small radius )

V (of small cylinder) = \pi(6.53)²(9)
                                  = 1206 in³
3 0
3 years ago
Ocean tides can be modeled by a sinusoidal function. Suppose that there is a low and high tide every 12 hours, and that high tid
USPshnik [31]

Answer:

(a)y=5sin(\frac{\pi}{6}(x+2))+5

(b) 7.5ft above the low tide.

Step-by-step explanation:

(a) To find the function that computes the height of the tide, you need to select the form of the sinusoidal function. For example, use the form:

y=Asin(B(x-C))+D

Where A is the amplitude, B the frequency, C the phase shift and D the vertical shift.

The  amplitude is half the distance between the highest and the lowest tide:

A=10/2=5ft

The frequency is related to the period T by:

B=\frac{2\pi}{T}

The period is 12 hours, then

B=\frac{2\pi}{12}=\frac{\pi}{6}

The high tide is at 1:00 a.m. and 1:00 p.m. , this is the moment when sin(B(x-C))=1, if sin(\frac{\pi}{2})=1 then B(x-C) must be equal to \frac{\pi}{2} when x=1:

B(x-C)=\frac{\pi}{2}\\\frac{\pi}{6}(1-C)=\frac{\pi}{2}\\\frac{1}{6}(1-C)=\frac{1}{2}\\(1-C)=\frac{6}{2}\\-C=3-1\\C=-2

The vertical shift is the sum of the lowest value, the height of the low tide (lt) and the amplitude:

D=5+lt

The function is:

y=5sin(\frac{\pi}{6} (x+2))+5+lt

Because the function must be the height above low tide height, subtract this heigh from the function:

y=5sin(\frac{\pi}{6} (x+2))+5+lt-lt

y=5sin(\frac{\pi}{6} (x+2))+5

(b) Use x=11 in the function

y=5sin(\frac{\pi}{6} (11+2))+5=2.5+5=7.5ft above the low tide.

3 0
4 years ago
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