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Musya8 [376]
2 years ago
11

Thomas has sixty-two cents. How much money is this? $0.62 $6.02 $62 $6.20 Done →

Mathematics
2 answers:
Anvisha [2.4K]2 years ago
3 0

Answer:

$0.62

Step-by-step explanation:

makvit [3.9K]2 years ago
3 0

Answer:

$0.62 ......................

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7. If a 12 pack of soda cost Mr. Key $3.84, what would the unit price be for
AlekseyPX

Answer:

32 cents

Step-by-step explanation:

3.84 ÷ 12 = 0.32 so 32 cents

7 0
3 years ago
Which terms in the following expression are like terms?
Alinara [238K]

9514 1404 393

Answer:

  (c)  5x and 3x, and 4 and 1

Step-by-step explanation:

Like terms have the same variable(s) to the same power(s).

The terms of this expression are ...

  • x^3: variable x, power 3
  • 5x: variable x, power 1
  • -3x: variable x, power 1
  • 3y: variable y, power 1
  • 4: no variable
  • -1: no variable

The like terms are {5x, -3x}, which have the x-variable to the first power, and {4, -1}, which have no variable.

6 0
2 years ago
Identifying Possible Triangles
Masja [62]

Answer:

3rd triangle can be constructed with dimensions 2,6,7.

Step-by-step explanation:

sum of any two sides > third side.

difference of any two sides < third side

1.

8+5=13 not >14 (no triangle.)

2.

7+8=15 not >15 (no triangle)

3.

2+6=8>7

2+7=9>6

7+6=13>2

7-2=5<6

7-6=1<2

6-2=4<7

so it is a triangle.

4.

6+3=9 not >10 (not a triangle)

5 0
2 years ago
What’s the answer show your WORK Plz
Yuliya22 [10]

Answer:

732.7 I think

Step-by-step explanation:

You just multiply 28 by 26.17

3 0
3 years ago
Read 2 more answers
1. y = x2 + 8x + 15<br> Find the zeros of the function by rewriting the function in intercept form
lubasha [3.4K]

The zeros of given function y=x^{2}+8 x+15 is – 5 and – 3

<u>Solution:</u>

\text { Given, equation is } y=x^{2}+8 x+15

We have to find the zeros of the function by rewriting the function in intercept form.

By using intercept form, we can put value of y as  to obtain zeros of function

We know that, intercept form of above equation is x^{2}+8 x+15=0

\text { Splitting } 8 x \text { as }(5+3) x \text { and } 15 \text { as } 5 \times 3

\begin{array}{l}{\rightarrow x^{2}+(5+3) x+5 \times 3=0} \\\\ {\rightarrow x^{2}+5 x+3 x+5 \times 3=0}\end{array}

Taking “x” as common from first two terms and “3” as common from last two terms

x (x + 5) + 3(x + 5) = 0

(x + 5)(x + 3) = 0

Equating to 0 we get,

x + 5 = 0 or x + 3 = 0

x = - 5 or – 3

Hence, the zeroes of the given function are – 5 and – 3

5 0
3 years ago
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