Ask question

Ask question

All categories

3 years ago

8

Please help giving brainiest

2 answers:

Klio2033 [76]3 years ago

3 0

I think it’s the last one sorry if I’m wrong

netineya [11]3 years ago

3 0

Answer:

I think so too is the last one

You might be interested in

Is the point (5, -7) a solution to the equation <br> Y > -2x + 3

nordsb [41]

No,isn’t solution
-7>-2*5+3
-7>-7 wrong

3 0

3 years ago

Simplify.<br> √20⋅√3·√18<br><br> 5√30<br><br> 5√10<br><br> 6√10<br><br> 6√30

ddd [48]

The correct answer is: [D]: " 6√30 " .

_____________________________________

Explanation:

_____________________________________

Simplify: " √20 ⋅ √3 · √18 " :

___________________________

Step 1) Simplify the "first term" :

"√20 " = √4 √5 = 2√5 ;

___________________________

Step 2): Consider the "second term:

"√3 " = √3 (already simplified);

___________________________

Step 3) Simplifly the "third term" :

" √18 " = √9 * √2 = " 3√2 " ;

____________________________

We have: " 2√5 * √3 * 3√2 " ;

= (2* 3) * (√5 * √3 * √2) ;

= 6 * (√5 √3 √2) ;

Note: " (√5 * √3 * √2 )" = " √(5 * 3 * 2) " = " √ 30 " .

So: " 6 * (√5 * √3 * √2) =

______________________________________

→ " 6 √30 " .

______________________________________

The answer is: " 6√30 " ;

→ which is: "Answer choice: [D]: " 6√30 " .

_____________________________________

Hope this answer helps!

Best wishes!

_____________________________________

5 0

3 years ago

Read 2 more answers

Let an be the sum of the first n positive odd integers.

sesenic [268]

Answer:

A) The first 4 terms of the sequences are: $a_{1} =16$ , $a_{2} =24$ , $a_{3} =32$ and $a_{4} =40$ .

B) An explicit formula for this sequence can be written as: $a_{n} =8*(n+1)$

C) A recursive formula for this sequence can be written as:

$\left \{ {{a_{1} =16} \atop {a_{n} =a_{n-1}+8}} \right.$

Step-by-step explanation:

A) You can find the firs terms of this sequence simply selecting an odd integer and summing the consecutive 3 ones:

$a_{n} = Odd_{n}+Odd_{n+1}+Odd_{n+2}+Odd_{n+3}$ (a.1)

$a_{1}=1+3+5+7=16$

$a_{2}=3+5+7+9=24$

$a_{3}=5+7+9+11=32$

$a_{4}=7+9+11+13=40$

B) Observe the sequence of odd numbers 1, 3, 5, 7, 9, 11, 13(...).

You can express this sequence as:

$Odd_{n}=(2*n-1)$ (b.1)

If you merge the expression b.1 in a.1, you obtain the explicit formula of the sequence:

$a_{n} = Odd_{n}+Odd_{n+1}+Odd_{n+2}+Odd_{n+3}$ (a.1)

$a_{n} = (2*n-1)+((2*(n+1)-1))+((2*(n+2)-1))+((2*(n+3)-1))$ (b.2)

$a_{n} = 8*n+8$ (b.3)

$a_{n} =8*(n+1)$ (b.s)

C) The recursive formula has to be written considering an initial term and an N term linked with the previous term. You can see an addition of 8 between a term and the next one. So you can express each term as an addition of 8 with the previous one. Therefore, if the first term is 16:

$\left \{ {{a_{1} =16} \atop {a_{n} =a_{n-1}+8}} \right.$ (c.s)

5 0

3 years ago

Tell which equation you would choose to solve for one of the variables. Explain.

lapo4ka [179]

Answer:

i would choose to solve x+2y=0 because it is simple and ez

Step-by-step explanation:

i took test and got 100 % brainliest plz

5 0

3 years ago

What is the correct first step to solve this system of equations by elimination?

storchak [24]

$\bold{\huge{\pink{\underline{ Solution }}}}$

$\bold{\underline{ Given }}$

<u>We </u><u>have </u><u>given </u><u>two </u><u>linear </u><u>equations </u><u>that</u><u> </u><u>is </u><u>2x </u><u>-</u><u> </u><u>3y </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>and </u><u>x</u><u> </u><u>+</u><u> </u><u>3y </u><u>=</u><u> </u><u>1</u><u>2</u><u> </u><u>.</u>

$\bold{\underline{ To \: Find }}$

<u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>x </u><u>and </u><u>y </u><u>by </u><u>elimination </u><u>method</u><u>. </u>

$\bold{\underline{ Let's \: Begin }}$

$\sf{ 2x - 3y = -6 ...eq(1)}$

$\sf{ x + 3y = 12 ...eq(2)}$

<u>Multiply </u><u>eq(</u><u> </u><u>2</u><u> </u><u>)</u><u> </u><u>by </u><u>2</u><u> </u><u>:</u><u>-</u>

$\sf{ 2( x + 3y = 12 )}$

$\sf{ 2x + 6y = 24 }$

<u>Subtract </u><u>eq(</u><u>1</u><u>)</u><u> </u><u>from </u><u>eq(</u><u>2</u><u>)</u><u> </u><u>:</u><u>-</u>

$\sf{ 2x + 6y -( 2x - 3y) = 24 -(-6)}$

$\sf{ 2x + 6y - 2x + 3y = 24 + 6 }$

$\sf{ 9y = 30 }$

$\sf{ y = 30/9}$

$\sf{\red{ y = 10/3}}$

<u>Now</u><u>, </u><u> </u><u>Subsitute</u><u> </u><u>the </u><u>value </u><u>of </u><u>y </u><u>in </u><u>eq(</u><u> </u><u>1</u><u> </u><u>)</u><u>:</u><u>-</u>

$\sf{ 2x - 3(10/3) = -6 }$

$\sf{ 2x - 10 = -6 }$

$\sf{ 2x = -6 + 10}$

$\sf{ x = 4/2}$

$\sf{\red{ x = 2}}$

Hence, The value of x and y is 2 and 10/3

6 0

3 years ago