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Angelina_Jolie [31]
3 years ago
10

Which graph contains an Euler path?

Mathematics
1 answer:
Otrada [13]3 years ago
6 0

Answer:

graph 1

Step-by-step explanation:

path = 2 odd degree vertices, the rest even

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I need help with this
Nostrana [21]

Answer:

28 times

Step-by-step explanation:

2.8414 x 10^9 ÷ 1.0127 x 10^8

(2.8414 ÷ 1.0127) · (10^9 ÷ 10^8)

2.8 x 10

28

4 0
3 years ago
Here is a quadratic equation. is it in Standard form? 2x^2-7x-3=0​
andrew-mc [135]

Answer:

Yes

Step-by-step explanation:

The standard form for a quadratic equation in the US has the terms in decreasing order by power on the left of the equal sign, and the right side zero. Preferably, the leading coefficient is positive. This might also be called "general form." Elsewhere, the "standard form" may be different.

6 0
3 years ago
(math)
slega [8]

Answer:

c rectangle I'm pretty sure

5 0
3 years ago
1.) A man invested $600 at 8% per annum for 5 years. Calculate: a.) The simple interest payable b.) The total amount of money th
Bezzdna [24]

Answer:

Following are the solution to this question:

Step-by-step explanation:

Using formula:

\text{Simple Interest} = \frac{P \times R \times T}{100} \\\\\text{ Total amount} = \text{principle}+  \text{Simple Interest}

In question (1):

Principle= \$ 600\\Rate = \ 8 \%\\Time= \ 5

In point a:

\to \frac{P \times R \times T}{100} \\\\\to \frac{(600 \times  8 \times 5)}{100}\\\\ \to  240

In point b:

\to amount = principle \ +\ simple \ interest

                 = 600 + 240 \\\\= 840

In question (2):

Principle= \$ 12,450\\Rate = \ 7.25 \%\\Time= \ 6

In point a:

\to \frac{P \times R \times T}{100} \\\\\to \frac{(12,450 \times  7.25 \times 6)}{100}\\\\ \to  5,415.75

In point b:

\to amount = principle \ +\ simple \ interest

                 = 12,450 + 5,415.75 \\\\= 17,865.75

The 3 question is incomplete, that's why it can't be solved.

7 0
3 years ago
Should Clayton buy this shower or not and why?
Artyom0805 [142]

Answer:

No he's too poor

Step-by-step explanation:

8 0
3 years ago
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