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Gelneren [198K]
3 years ago
13

Robert can walk 5/8 mile in 1/2. What was his average speed in miles per hour

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
7 0

Answer:

The answer is 1.23 miles per hour

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Select all the points that are on the graph of the line
Stolb23 [73]

Answer:

(0,-3), (-5,-5)

Step-by-step explanation:

Plug in x and y and see in which case = -15

-2(0)+5(5)= 25 no

-2(0)+5(-3)= -15 yes

-2(-5)+5(-1)= 5 no

-2(-5)+5(-5)= -15 yes

-2(10)+5(-1)= -25 no

-2(-10)+5(7)= 55 no

4 0
2 years ago
Help please asap
padilas [110]
These are 2 questions and 2 answers:

<span>1. Question 1: The area of a rectangular carpet is given by the trinomial 5x^2 - 3x - 14. What are the possible dimensions of the carpet? Use factoring.

Answer: 5x^2 -3x  -14

Explanation:

You might probe the different pairs of factors given in the answer choices to check wich product is equal to the oirginal polynomial or you can factor from zero.

I will show  you how to factor that polynomial from zero:

1) given: 5x^2 - 3x - 14

2) multiply and divide by the leading coefficient =>

5 [5x^2 - 3x  -14 ]
----------------------
         5

3) Enter the the factor 5 inside the parenthesis:

5.5x^2 - 5.3x - 5.14
--------------------------
              5

4) rearrange conviniently to show (5x) as a common factir of the first two terms

(5x)^2 - 3(5x) - 70
--------------------------
              5

5) factor the numerator using 5x as common term of the two binomials:

(5x  ) ( 5x   ) <----> open the parenthesis
(5x - ) (5x + ) <-----> include the signs
(5x - 10) (5x + 7)  <---> two numbers that add up - 3 and their product is  - 70

=>

(5x - 10)(5x + 7)
---------------------
         5

6) divide (5x - 10) by 5 => x - 2

7) factored expression

(x - 2) (5x + 7)

8) Conclusion: the answer is the option B. (5x + 7) and (x - 2)


2. Question 2.  The area of a rectangular barnyard is given by the trinomial 6x^2 + 7x – 20. What are the possible dimensions of the barnyard? Use factoring

Answer: </span><span><span>option C. (2x + 5)(3x - 4)</span>

Explanation:

The procedure is the same of the question 1.

1) given: 6x^2 + 7x - 20

2) multiply numerator and denominator by 6 and rearrange to show (6x) as common factor ot the first two terms =>

(6x)^2 + 7(6x) - 120
--------------------------
              6
3) factor the numerador using 6x as common factor ot two binomials, and simplify:

(6x + )(6x - )
(6x + )(6x - )
(6x + 15) (6x - 8)

=>

(6x + 15)(6x -8)
---------------------
           6

(6x + 15)(6x - 8)
---------------------- =
        3.2

(6x + 15)      (6x - 8)
------------- .  ------------ =
      3                2

(2x + 5)(3x - 4)

4) answer: option C. (2x + 5)(3x - 4)</span>
6 0
3 years ago
1. Ali's health club membership costs $19.95 per
shusha [124]

Answer:

10 visits tops

Step-by-step explanation:

50- 19.95= 30.05

30.05/2.75= 10.9 which means 10 times.

8 0
3 years ago
A cylinder has a volume 1200 cubic inches and radius 5 inches. What is the height of the cone to the nearest tenth inch
mihalych1998 [28]

Answer: 15.3 in

Step-by-step explanation:

First, the formula for the volume is: V=\pi r^2h

If you solve for h,

h=\frac{V}{\pi r^2}

h=\frac{1200in^3}{(3.14)(5in)^2}

h=15.3in

3 0
3 years ago
If the heights of 300 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many studen
Lunna [17]
Given:
μ = 68 in, population mean
σ = 3 in, population standard deviation

Calculate z-scores for the following random variable and determine their probabilities from standard tables.

x = 72 in:
z = (x-μ)/σ = (72-68)/3 = 1.333
P(x) = 0.9088

x = 64 in:
z = (64 -38)/3 = -1.333
P(x) = 0.0912

x = 65 in
z = (65 - 68)/3 = -1
P(x) = 0.1587

x = 71:
z = (71-68)/3 = 1
P(x) = 0.8413

Part (a)
For x > 72 in, obtain
300 - 300*0.9088 = 27.36

Answer: 27

Part (b)
For x ≤ 64 in, obtain
300*0.0912 = 27.36

Answer: 27

Part (c)
For 65 ≤ x ≤ 71, obtain
300*(0.8413 - 0.1587) = 204.78

Answer: 204

Part (d)
For x = 68 in, obtain
z = 0
P(x) = 0.5
The number of students is
300*0.5 = 150

Answer: 150

3 0
3 years ago
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