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Triss [41]
3 years ago
14

Help please I’ll mark you brainliest

Mathematics
1 answer:
goblinko [34]3 years ago
7 0
\mathrm{Compute\:the\:ratios\:of\:all\:the\:adjacent\:terms}:\:r=\frac{a_{n+1}}{a_n}

\frac{36}{6}=6,\:\quad \frac{216}{36}=6,\:\quad \frac{1296}{216}=6,\:\quad \frac{7776}{1296}=6

\mathrm{The\:ratio\:of\:all\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to} \ \textgreater \  r = 6

There is a bit more to the answer but it just explains a bit more that this question does not ask for.

Hope this helps!
You might be interested in
Find an odd number with 3 digits such that all the digits are different and add up to 15. The difference between the first two d
Serjik [45]

Answer:

The three digit number = 951

Step-by-step explanation:

Let suppose the numbers are:

= abc

According to given condition:

a+ b + c = 15  -------------eq1

Also, given the difference between the first two digit = the difference between the last two digits:  

==> l a-b l = l b-cl  

==> (a-b) = (b-c)  

==> (a+c) = 2b

Now we will substitue in eq1

==> a+ b + c = 15

==> 2b + b = 15

==> 3b = 15

Dividing both sides by 3 we get:

b =5

a + c = 2b

a+ c = 10

a = 10 -c ..........(2)

We know that"  

(a-b) = (b-c)

==> a > b+c

==> a > 5 + c

==> 10 -c > 5 +c  

==>5 > 2c

==> 2.5 > c

As c is an odd number so c will be equal to 1

c = 1

a = 10 -1

a = 9

The three digit number = 951

The hundred digit is greater than the sum of the tens and ones digits

i hope it will help you!

6 0
3 years ago
Which statement correctly compares the numbers 269,317 and 296,317?
MrRissso [65]

Your answer is c, because 69,000 is less than 96,000. I hope this helped! Could I possibly get brainliest?

5 0
2 years ago
Read 2 more answers
High school Geometry
xxTIMURxx [149]

Answer:

the answer is b. sss is not possible

6 0
2 years ago
A bird in the sky flies down.
tino4ka555 [31]

Answer:

Step-by-step explanation:

What is the rest of the equation? I will edit/fix the answer after you finish the equation.

5 0
3 years ago
Please help with math and please show your work will give brainliest
Burka [1]

Answer:

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d,

if  and  then a=bc=da+c=b+d

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

3x+y=52x−y=0–––––––––––5x=5

The y’s add to zero and we have one equation with one variable.

Let’s try another one:

{x+4y=22x+5y=−2(5.3.3)

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2, we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2.

This figure shows two equations. The first is negative 2 times x plus 4y in parentheses equals negative 2 times 2. The second is 2x + 5y = negative 2. This figure shows two equations. The first is negative 2x minus 8y = negative 4. The second is 2x + 5y = -negative 2.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Add the equations yourself—the result should be −3y = −6. And that looks easy to solve, doesn’t it? Here is what it would look like.

This figure shows two equations being added together. The first is negative 2x – 8y = −4 and 2x plus 5y = negative 2. The answer is negative 3y = negative 6.

We’ll do one more:

{4x−3y=103x+5y=−7

It doesn’t appear that we can get the coefficients of one variable to be opposites by multiplying one of the equations by a constant, unless we use fractions. So instead, we’ll have to multiply both equations by a constant.

We can make the coefficients of x be opposites if we multiply the first equation by 3 and the second by −4, so we get 12x and −12x.

This figure shows two equations. The first is 3 times 4x minus 3y in parentheses equals 3 times 10. The second is negative 4 times 3x plus 5y in parentheses equals negative 4 times negative 7.

This gives us these two new equations:

{12x−9y−12x−20y=30=28

When we add these equations,

\[{12x−9y=30−12x−20y=28–––––––––––––––––−29y=58

\]

the x’s are eliminated and we just have −29y = 58.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution

Step-by-step explanation:

Here is some examples

4 0
2 years ago
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