Translate eighty-eight less than m is smaller than −176

Mathematics

2 answers:

xeze [42]2 years ago

6 0

Answer:

$88$

Step-by-step explanation:

Katarina [22]2 years ago

3 0

Answer:

88 < m < -176

You might be interested in

An inventory record for the bookstore of a local college shows a receipt of 150 accounting books on May 1. Another shipment of 2

babunello [35]

Hi there
Let's order the transactions
on May 1. 150
less On May 20 100 books were sold

Less June 3, 31 books were purchased by the students

Add on June 15 25 books arrived

Add On June 17, a shipment of 150 books arrived

So
the number on hand on June 17
150−100−31+25+150
=194....answer

Hope it helps

4 0

3 years ago

Find the length UV A 9 units B about 3.0 C 65 units D about 8.1 units

wariber [46]

Answer:

D. I got it right on my quiz.

Step-by-step explanation:

5 0

2 years ago

Please help with this math question :)

Nikitich [7]

I can't see the math

5 0

3 years ago

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

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