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GarryVolchara [31]
3 years ago
5

Urgent, please help, I really appreciate (100 points)

Mathematics
1 answer:
ANTONII [103]3 years ago
6 0

Answer:

3, 6, 9, 12 is not geometric

Step-by-step explanation:

A geometric progression has a common ratio r between consecutive terms.

3, 6, 9 , 12

has a common difference of 3 between terms and is arithmetic

1, 5, 25, 125

r = 5 ÷ 1 = 25 ÷ 5 = 125 ÷ 25 = 5 ← geometric

4, 8, 16, 32

r = 8 ÷ 4 = 16 ÷ 8 = 32 ÷ 16 = 2 ← geometric

2, 6, 18, 54

r = 6 ÷ 2 = 18 ÷ 6 = 54 ÷ 18 = 3 ← geometric

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Find the area of the shape below
Westkost [7]

Answer:

46.875

Step-by-step explanation:

A= l x w x h so you multiple all three numbers together and get your answer

3 0
3 years ago
James drove up 6,400 feet to a camp area on a mountain to meet his friends. They walked a trail that led to the mountain summit
gregori [183]
The answer is 4900 because you add the 6,400+1,800=8,200 then you subtract the 8,200-3,300= 4,900
6 0
3 years ago
Read 2 more answers
What is the coefficient of x2y3 in the expansion of (2x + y)5?
Zigmanuir [339]

Option C:

The coefficient of x^{2} y^{3} is 40.

Solution:

Given expression:

(2 x+y)^{5}

Using binomial theorem:

(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}

Here a=2 x, b=y

Substitute in the binomial formula, we get

(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}

                                                            $+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}

Let us solve the term one by one.

$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}

$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y

$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}

$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}

$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}

$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}

Substitute these into the above expansion.

(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}

The coefficient of x^{2} y^{3} is 40.

Option C is the correct answer.

5 0
3 years ago
The algebra tiles represent the perfect square trinomial x2 10x c. what is the value of c? c =
devlian [24]

The value of c for which the considered trinomial becomes perfect square trinomial is: 20 or -20

<h3>What are perfect squares trinomials?</h3>

They are those expressions which are found by squaring binomial expressions.

Since the given trinomials are with degree 2, thus, if they are perfect square, the binomial which was used to make them must be linear.

Let the binomial term was ax + b(a linear expression is always writable in this form where a and b are constants and m is a variable), then we will obtain:

(ax + b)^2 = a^2x^2 + b^2 + 2abx

Comparing this expression with the expression we're provided with:

x^2 + 10x + c

we see that:

a^2 = 1 \implies a = \pm 1\\b^2 = 10\\b = \pm 10\\2ab = c\\\pm2(10)1 = c\\c = \pm 20\\

Thus, the value of c for which the considered trinomial becomes perfect square trinomial is: 20 or -20

Learn more about perfect square trinomials here:

brainly.com/question/88561

6 0
2 years ago
What is the range of the function f(x) = 3x - 5, for the domain { -1, 2, 4}
morpeh [17]
Hope this helps!! Solved it step by step:)

7 0
2 years ago
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