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WITCHER [35]
3 years ago
6

The relationships for Paul's and Steve's height with respect to Theresa's height are described above. First you will write expre

ssions for Paul's height and for Steve's height. Then you will relate the expressions.
Mathematics
2 answers:
Varvara68 [4.7K]3 years ago
7 0

Answer:

Theresa’s height is t inches. Paul says he is 16 inches shorter than 1 1/2 times Theresa’s height.

Paul's height in inches

= 1 1/2⁢t−16

= 3/2⁢t−16

The expression that represents Paul’s height in inches is 3/2t−16.

Theresa’s height is t inches. Steve says he is 6 inches shorter than 1 1/3 times Theresa’s height.

Paul's height in inches

= 1 1/3⁢t−6  

= 4/3t−6

The expression that represents Steve’s height in inches is 4/3⁢t−6.

Step-by-step explanation:

Makovka662 [10]3 years ago
5 0

Answer:

They are tall, im not sure

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which of the functions have a range of real numbers greater than or equal to 1 or less than or equal to-1​
lilavasa [31]

The question is incomplete. Here is the complete question:

Which of the functions have a range of all real numbers greater than or equal to 1 or less than or equal to -1? check all that apply.

A. y=\sec x

B. y= \tan x

C. y= \cot x

D. y= \csc x

Answer:

A. y=\sec x

D. y=\csc x

Step-by-step explanation:

Given:

The range is greater than or equal to 1 or less than or equal to -1.

The given choices are:

Choice A: y=\sec x

We know that, the \sec x=\frac{1}{\cos x}

The range of \cos x is from -1 to 1 given as [-1, 1]. So,

|\cos x|\leq 1\\\textrm{Taking reciprocal, the inequality sign changes}\\\frac{1}{|\cos x|}\geq 1\\|\sec x|\geq 1

Therefore, on removing the absolute sign, we rewrite the secant function as:

\sec x\leq -1\ or\ \sec x\geq 1\\

Therefore, the range of y=\sec x is all real numbers greater than or equal to 1 or less than or equal to-1​.

Choice B: y= \tan x

We know that, the range of tangent function is all real numbers. So, choice B is incorrect.

Choice C: y= \cot x

We know that, the range of cotangent function is all real numbers. So, choice C is incorrect.

Choice D: y=\csc x

We know that, the \csc x=\frac{1}{\sin x}

The range of \sin x is from -1 to 1 given as [-1, 1]. So,

|\sin x|\leq 1\\\textrm{Taking reciprocal, the inequality sign changes}\\\frac{1}{|\sin x|}\geq 1\\|\csc x|\geq 1

Therefore, on removing the absolute sign, we rewrite the cosecant function as:

\csc x\leq -1\ or\ \csc x\geq 1\\

Therefore, the range of y=\csc x is all real numbers greater than or equal to 1 or less than or equal to-1​.

6 0
3 years ago
ANDREA IS A VERY GOOD STUDENT. THE PROBABILITY THAT SHE STUDIES AND PASSES HER MATHEMATICS TEST IS 17/20. IF THE PROBABILITY THA
Eva8 [605]
Let B be the event that Andrea passes her test, and let A be the event 
<span>that she studies. We are given that P(A and B) = 17/20, and that P(A) = 15/16. </span>

<span>Now the probability that Andrea passes her test given that she has studied </span>
<span>is represented by P(BlA). The formula your teacher gave you can be written as </span>
<span>P(BlA) = P(A and B) / P(A). </span>

<span>So P(BlA) = P(A and B) / P(A) = (17/20) / (15/16) = (17/20)*(16/15) = 68/75.</span>
3 0
3 years ago
Read 2 more answers
A rectangular room is 3 meters longer than it is wide, and its perimeter is 22 meters. Find the dimension of the room. The lengt
lyudmila [28]

Answer:

Width = 4 m

Length = 7 m

Step-by-step explanation:

given:

perimeter of a rectangle = 22m

L = 3 + W

perimeter = 2L + 2W

perimeter = 2 (3 + W) + 2W

22 = 6 + 2W + 2W

22 - 6 = 4W

W = 16 / 4

W = 4 m

L = 3 + W

L = 3 + 4

L = 7 m

check:

perimeter = 2L + 2W

22 = 2(7) + 2(4)

22 = 14 + 8

22 = 22  ---- OK

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The price of a cell phone can be modeled by the equation:
Igoryamba
4 years I’m pretty sure.
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GarryVolchara [31]

Step-by-step explanation:

1. 1/4(x + 12)=2 - first isolate X'

         - 12 -12

2. 1/4(x) = -10  - 1/4  since its a fraction, you'll have to subtract not divide

-1/4    - 1/4

3. x = -10.25

Andre did the first step correctly, but on the second step he just divided 1.4 by -10, instead of subtract 1/4 from both sides.

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