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3 years ago

13

Will give brainliest!!

2 answers:

kirill115 [55]3 years ago

8 0

It’s the one that is marked as (3,3)

blagie [28]3 years ago

4 0

Answer:

Second one

Step-by-step explanation:

The first line has a negative slope, that is means that it decreases (we have to watch the graph from left to right)

6 is the point where it Intercepts the y axis

For now we can say that only the second and the fourth graph can be correct

The second line has a positive slope, so it increases (we have to watch the graph from left to right)

-3 is the point where it intercepts the y axis

the second graph shows what we want

You might be interested in

Please help explanation if possible

Eva8 [605]

Answer:

612 adults

361 students

Step-by-step explanation:

To solve this question, set two equations:

Let x be number of adults and y be number of students.

As there are in total 937 people, the equation would be the sum of both adults and children:

$x+y=937$

$x=937-y$ ...... ( 1 )

As the total sale amount is $1109, the equation would be to add up the ticket fee:

$2x+0.75y=1,109$ ...... ( 2 )

Put ( 1 ) into ( 2 ):

$2(937-y)+0.75y=1,109$

$1874-2y+0.75y=1,109$

$-1.25y=-765$

$y=763/1.25$

$y=612$

Put y into ( 1 ):

$x=973-612$

$x=361$

Therefore there are 612 adults and 361 students.

5 0

2 years ago

Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$

Viktor [21]

When $n=0$ , we have

$5^{3^0} + 1 = 5^1 + 1 = 6$

$3^{0 + 1} = 3^1 = 3$

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for $n=k$ , that

$3^{k + 1} \mid 5^{3^k} + 1$

Now for $n=k+1$ , we have

$5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)$

so we know the left side is at least divisible by $3^{k+1}$ by our assumption.

It remains to show that

$3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1$

which is easily done with Fermat's little theorem. It says

$a^p \equiv a \pmod p$

where $p$ is prime and $a$ is any integer. Then for any positive integer $x$ ,

$5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3$

Furthermore,

$5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3$

which goes all the way down to

$5^{3^k} \equiv 5 \pmod 3$

So, we find that

$\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3$

QED

5 0

1 year ago

Which expression represents: twice a number increased by 39

Cerrena [4.2K]

Answer:

2x+39

Step-by-step explanation:

8 0

2 years ago

Help please!!! Best answer will get brainliest!!!

r-ruslan [8.4K]

Answer is b for this question

3 0

2 years ago

What are the leading coefficient, constant term and degree, if any, of the algebraic expression

nasty-shy [4]

Coefficients are the numbers in from with the variable, the constant is just a number and the degree is the exponent.

6 0

3 years ago