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son4ous [18]
1 year ago
12

A cable company claims that the average household pays $78 a month for a basic cable plan, but it could differ by as much as $20

. Write an absolute value inequality to determine the range of basic cable plan costs with this cable company.
A. |x − 78| ≥ 20
B. |x − 20| ≥ 78
C. |x − 20| ≤ 78
D. |x − 78| ≤ 20
Mathematics
2 answers:
Marrrta [24]1 year ago
8 0
Answer is B cause you have to minus the 20 then then swap the signs
Margarita [4]1 year ago
7 0

Answer:

D. |x − 78| ≤ 20

Step-by-step explanation:

Given,

The monthly charges for a basic cable plan = $ 78,

Also,  it could differ by as much as $20,

So, the maximum charges = $(78 + 20) ,

And, the minimum charges = $(78 - 20),

Let x represents the monthly charges ( in dollars ),

78 - 20 ≤ x ≤ 78 + 20

⇒ 78 - 20 ≤ x and x ≤ 78 + 20

⇒ -20 ≤ x -78 and x-78 ≤ 20

⇒ 20 ≥ -(x-78) and x-78 ≤ 20   ( ∵ a > b ⇒ -a < -b )

⇒ |x-78| ≤ 20

Which is the required absolute value inequality to determine the range of basic cable plan costs,

Option 'D' is correct.

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In total, she will use

$251.99 + $150.00 + $30.98 = $432.97.

If she has $840.32 in savings, then

$840.32 - $432.97 = $407.35 will remain.

Answer: she will use $432.97 - correct answer A

5 0
2 years ago
Given that line s is perpendicular to line t, which statements must be true of the two lines? Check all that apply.
artcher [175]
The basis to respond this question are:

1) Perpedicular lines form a 90° angle between them.

2) The product of the slopes of two any perpendicular lines is - 1.

So, from that basic knowledge you can analyze each option:

<span>a.Lines s and t have slopes that are opposite reciprocals.

TRUE. Tha comes the number 2 basic condition for the perpendicular lines.

slope_1 * slope_2 = - 1 => slope_1 = - 1 / slope_2, which is what opposite reciprocals means.

b.Lines s and t have the same slope.

FALSE. We have already stated the the slopes are opposite reciprocals.

c.The product of the slopes of s and t is equal to -1

TRUE: that is one of the basic statements that you need to know and handle.

d.The lines have the same steepness.

FALSE: the slope is a measure of steepness, so they have different steepness.

e.The lines have different y intercepts.

FALSE: the y intercepts may be equal or different. For example y = x + 2 and y = -x + 2 are perpendicular and both have the same y intercept, 2.

f.The lines never intersect.

FALSE: perpendicular lines always intersept (in a 90° angle).

g.The intersection of s and t forms right angle.

TRUE: right angle = 90°.

h.If the slope of s is 6, the slope of t is -6

FALSE. - 6 is not the opposite reciprocal of 6. The opposite reciprocal of 6 is - 1/6.

So, the right choices are a, c and g.
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5 0
1 year ago
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Which of the following best describes the figure?
stira [4]

Answer:

a convex nonagon

Step-by-step explanation:


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1 year ago
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Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |
Bas_tet [7]

Answer:

M=168k

(\bar{x},\bar{y})=(5,\frac{85}{28})

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

M=\int\int \rho dA

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx

Let's solve this integral:

M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx

M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx      

M=k\int^{9}_{1}21dx

M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}

So the mass will be:

M=21k*8=168k

Now we need to find the x-coordinate of the center of mass.

\bar{x}=\frac{1}{M}\int\int x*\rho dydx

\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx

\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx

\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx

\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx

\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}

\bar{x}=\frac{21}{168}*40=5

Now we need to find the y-coordinate of the center of mass.

\bar{y}=\frac{1}{M}\int\int y*\rho dydx

\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx

\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx

\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx

\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx

\bar{y}=\frac{255}{672}\int^{9}_{1}dx

\bar{y}=\frac{255}{672}8=\frac{2040}{672}

\bar{y}=\frac{85}{28}

Therefore the center of mass is:

(\bar{x},\bar{y})=(5,\frac{85}{28})

I hope it helps you!

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2 years ago
There were 567 people at a concert when a band started playing. After each song, only one-third of the people stayed to hear the
Solnce55 [7]
To write the function correctly, it is important to assign variables correctly and understand the situation of the problem clearly. For this, we let y the number of people and x as the number of songs played.

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at x = 2   y = 567 - 567(1/3)(1/3)
at x = 3   y = 567 - 567(1/3)(1/3)(1/3)

Therefore, the number of people left after x songs would be represented by the equation:

y = 567 - 567(1/3)x
y = 567 ( 1- x/3 )
7 0
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