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Ratling [72]
3 years ago
7

Which of the following numbers is a square number? Check all that apply.

Mathematics
1 answer:
svetoff [14.1K]3 years ago
6 0
A and D are square numbers.
You might be interested in
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
Suppose you have 13 tubes of paint. How many distinct color groupings can you make with your paint?
anzhelika [568]

Answer:

32767 distinct color grouping

Step-by-step explanation:

Given data:

Total tube is 13

for distinct color, group can be 1,2,3,4,.... 15

For total number of ways we have following relation

^nC_0 +^nC_1 +^nC_2 + .......^nC_n = 2^n

here n = 13 so we have

^15C_0 +^15C_1 +^15C_2 + .......^15C_{15} = 2^15

for at least one

= 2^{15} - 1

= 32768 -1

= 32767 distinct color grouping

3 0
3 years ago
Geometry (19)<br><br> Study Guide <br><br> My answer: x = 11 so 2(11) - 6 = 16, FG = 16
jok3333 [9.3K]
I think you meant to say n = 11. If so, then you are correct and it leads to the proper length for FG, which is 16 units. Nice work. 
3 0
4 years ago
Help me and thx alot
Svet_ta [14]

Answer:

easyyyyyyy

Step-by-step explanation:

table J!

8 0
3 years ago
Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant
Viktor [21]

Answer:

Therefore k= \frac{ln2 }{18}, A=184

Step-by-step explanation:

Given function is

T(t)=230 -e^{-kt}

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

138=230 -Ae^{-k\times 18}

\Rightarrow  -Ae^{-18k}=138-230

\Rightarrow  Ae^{-18k}=92 .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

184=230-Ae^{-k\times 36}

\Rightarrow Ae^{-36k}=230-184

\Rightarrow Ae^{-36k}=46.......(2)

Dividing (2) by (1)

\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}

\Rightarrow e^{-18k}=\frac{46}{92}

Taking ln both sides

ln e^{-18k}=ln\frac{46}{92}

\Rightarrow -18k =ln (\frac12)

\Rightarrow -18k= ln1-ln2

\Rightarrow k= \frac{ln2 }{18}

Putting the value k in equation (1)

Ae^{-18\frac{ln2}{18}}=92

\Rightarrow A e^{ln2^{-1}}=92

\Rightarrow A.2^{-1}=92

\Rightarrow \frac{A}{2}=92

\Rightarrow A= 92 \times 2

⇒A= 184.

Therefore k= \frac{ln2 }{18}, A=184

7 0
3 years ago
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