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barxatty [35]
3 years ago
8

Evaluate for a = 2 5a + 6 = _____

Mathematics
2 answers:
Aleksandr [31]3 years ago
7 0
5(2)+6

Pemdas

So 5(2)=10

Then add

10+6=16
aleksandrvk [35]3 years ago
5 0
Since a=2 it will be 5x2 which equal 10, plus 6
You might be interested in
Subtract 4 1/5 - 3/4
GaryK [48]

Change both fractions, so they have the same denomater

4 1/5= 4 4/20

3/4= 15/20

change the mixed number into an improper fraction

84/20Subtract 15 from 84

84-15=69

so you get 69/20

Hope this helped :)

7 0
3 years ago
I need help answering the question in the picture provided
adoni [48]

answer:

(a).

Equation of circle is <u>x²</u><u> </u><u>+</u><u> </u><u>y²</u><u> </u><u>-</u><u> </u><u>2</u><u>5</u><u> </u><u>=</u><u> </u><u>0</u>

(b).

(-5, 0) » yes

(√7, 1) » no

(-3, √21) » no

(0, 7) » no

Step-by-step explanation:

(a).

If centred at origin, centre is (0, 0)

General equation of circle:

{ \boxed{ \bf{ {x}^{2} +  {y}^{2} + 2gx + 2fy   + c = 0}}}

but g and f are 0:

{x}^{2}  +  {y}^{2}   + c = 0 \\ but :  \\ c =  {g}^{2} +  {f}^{2}  -  {r}^{2}  \\ c =  { - 25}

8 0
3 years ago
Some people think it is unlucky if the 13th day of month falls on a Friday. show that in that there year (non-leap or leap) ther
Vlad1618 [11]
<span>There are several ways to do this problem. One of them is to realize that there's only 14 possible calendars for any year (a year may start on any of 7 days, and a year may be either a leap year, or a non-leap year. So 7*2 = 14 possible calendars for any year). And since there's only 14 different possibilities, it's quite easy to perform an exhaustive search to prove that any year has between 1 and 3 Friday the 13ths. Let's first deal with non-leap years. Initially, I'll determine what day of the week the 13th falls for each month for a year that starts on Sunday. Jan - Friday Feb - Monday Mar - Monday Apr - Thursday May - Saturday Jun - Tuesday Jul - Thursday Aug - Sunday Sep - Wednesday Oct - Friday Nov - Monday Dec - Wednesday Now let's count how many times for each weekday, the 13th falls there. Sunday - 1 Monday - 3 Tuesday - 1 Wednesday - 2 Thursday - 2 Friday - 2 Saturday - 1 The key thing to notice is that there is that the number of times the 13th falls upon a weekday is always in the range of 1 to 3 days. And if the non-leap year were to start on any other day of the week, the numbers would simply rotate to the next days. The above list is generated for a year where January 1st falls on a Sunday. If instead it were to fall on a Monday, then the value above for Sunday would be the value for Monday. The value above for Monday would be the value for Tuesday, etc. So we've handled all possible non-leap years. Let's do that again for a leap year starting on a Sunday. We get: Jan - Friday Feb - Monday Mar - Tuesday Apr - Friday May - Sunday Jun - Wednesday Jul - Friday Aug - Monday Sep - Thursday Oct - Saturday Nov - Tuesday Dec - Thursday And the weekday totals are: Sunday - 1 Monday - 2 Tuesday - 2 Wednesday - 1 Thursday - 2 Friday - 3 Saturday - 1 And once again, for every weekday, the total is between 1 and 3. And the same argument applies for every leap year. And since we've covered both leap and non-leap years. Then we've demonstrated that for every possible year, Friday the 13th will happen at least once, and no more than 3 times.</span>
5 0
3 years ago
A rectangular prism has a volume of 729 ft'. The length, width, and height are all the same. What is the length of each side of
german

Answer:

9

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Which expressions represents twelve diminished by six times a number:. 12-6n; 6n-12; 12n-6
Ede4ka [16]
12-6n
.......................
4 0
3 years ago
Read 2 more answers
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