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yan [13]
3 years ago
12

find 3 consecutive even integers such that the sum of the smaller 2 is 6 less than twice the largest one

Mathematics
1 answer:
saw5 [17]3 years ago
6 0

n, n + 2, n + 4 - three consecutive even integers

n + 2 = 2(n + 4) - 6     |use distributive property

n + 2 = (2)(n) + (2)(4) - 6

n + 2 = 2n + 8 - 6

n + 2 = 2n + 2     |subtract 2 from both sides

n = 2n       |subtract 2n from both sides

-n = 0

n = 0

n + 2 = 0 + 2 = 2

n + 4 = 0 + 4 = 4

Answer: 0, 2, 4.

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I need help with finding out both of the answers to these questions.
Semmy [17]

Answer:

6). unknown side is 6.23m

7). angle is 26.58 degrees

Step-by-step explanation:

you must use the Pythagoras theorem formula to find the unknown side if you have only two sides

7 0
3 years ago
Read 2 more answers
1. A sine function has the following key features:
andrew11 [14]
Problem 1

See the attached image (figure 1)

16pi seems like a typo. I'm going to assume that it's a fraction and it is 1/(6pi)
f = 1/(6pi) = frequency
T = 1/f = 1/(1/(6pi)) = 6pi
Amplitude = 2
a = 2
b = 2pi/T = 2pi/(6pi) = 1/3
Midline: y = 3
d = 3

The function is
y = a*sin(bx-c)+d
y = 2*sin(1/3*x-0)+3
y = 2*sin(x/3)+3

===============================================

Problem 2

See the attached image (figure 2) 

T = 12 is the period
a = 4 is the amplitude
b = 2pi/T = 2pi/12 = pi/6
y = 1 is the midline so d = 1
The y intercept is (0,1) which is the midline, which indicates no phase shifts have occurred so c = 0

The function is
y = a*sin(bx-c)+d
y = 4*sin((pi/6)x-0)+1
y = 4*sin((pi/6)x)+1

===============================================

Problem 3

See the attached image (figure 3)

Period = 4pi
T = 4pi
b = 2pi/T = 2pi/(4pi) = 1/2 = 0.5
Amplitude = 2
a = 2
Midline: y = 3
d = 3
y-intercept: (0,3)
The function is a reflection of its parent function over the x-axis, so 'a' is negative meaning a = -2 instead of a = 2

The function is
y = a*sin(bx-c)+d
y = -2*sin(0.5x-0)+3
y = -2*sin(0.5x)+3

===============================================

Problem 4

See the attached image (figure 4)

a = 10 which is half of the distance between the highest and lowest points
T = 8 is the period
b = 2pi/T = 2pi/8 = pi/4
c = -pi/2 is the phase shift since its really a cosine graph
d = 0 is the midline

The function is
y = a*sin(bx-c)+d
y = 10*sin((pi/4)*x+(-pi/2))+0
y = 10*sin((pi/4)*x+pi/2)

===============================================

Problem 5

See the attached image (figure 5)

a = 2 is the amplitude since it bobs up and down this distance from the midline
T = 8 seconds is the period (double that of the time it takes for it to go from the highest to the lowest point)
b = 2pi/T = 2pi/8 = pi/4
c = 0 is the phase shift as the buoy starts at normal depth of 20 meters
d = 20 is the midline

The function is
y = a*sin(bx-c)+d
y = 2*sin((pi/4)x-0)+20
y = 2*sin((pi/4)x)+20

===============================================

8 0
2 years ago
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Translate the following statement into a mathematical equation:
sleet_krkn [62]
(6+4)n =18 is the answer
8 0
2 years ago
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Part of a hundredths grid is shown.
olga_2 [115]

Answer:

a =  6/5

b = 0.111 . . .

c =  166 2/3%

Step-by-step explanation:

just did it

4 0
3 years ago
Find NQ if NP = 13 cm and line l is a segment bisector.
Zielflug [23.3K]

Answer:

NQ = 26 cm

Step-by-step explanation:

A line bisector divides a line into two equal segments.

Given that line l is the bisector of segment NQ, it then means it divides NQ into two equal segments, namely, segment NP and segment PQ.

Thus, NP = ½ of segment NQ

Therefore, if NP = 13 cm,

13 = ½*NQ

multiply both sides by 2

2*13 = NQ

26 = NQ

NQ = 26 cm

7 0
3 years ago
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