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victus00 [196]
3 years ago
13

List the first 5 mutiplies of 6 9 and 7

Mathematics
2 answers:
NeX [460]3 years ago
7 0
For 6: 12, 18, 24, 30, 36
For 9: 18, 27, 36, 45, 54
For 7: 14, 21, 28, 35, 42
Nat2105 [25]3 years ago
7 0
6, 12, 18, 24, 30, 36
9, 18, 27, 36, 45, 54
7, 14, 21, 28, 35, 42
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Please help me ASAP!!! I will give 20 points..... bad answers will be reported
belka [17]
150? I think thats the anwser
3 0
3 years ago
If your starting salary is $40,000 and you  receive a 4.2% raise every year, how much  TOTAL money will you make in 15 years?
Oliga [24]
The first year, the amount is 40,000

the second year is 40000 + 4.2% of 40000, or 0.042 * 4000, so 40000+(0.042*4000)
common factoring that  we get 40000(1 + 0.042), or just 40000(1.042)

in short, the starting amount is 40000, and to get the next term's value you'd use the "common ratio" of 1.042, namely the multiplier of 1.042.

for the third year it'll be 40000(1.042) + (0.042 *40000(1.042) ), again, common factoring that

40000(1.042)(1 + 0.042) or 40000(1.042)(1.042) or 40000(1.042)²

therefore,

\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
a_1=40000\\
r=1.042\\
n=15
\end{cases}
\\\\\\
S_{15}=40000\left( \cfrac{1-1.042^{15}}{1-1.042} \right)\implies S_{15}\approx 40000\left( \cfrac{-0.8536}{-0.042} \right)
\\\\\\
S_{15}\approx 812951.42
5 0
3 years ago
Find the x <br> please will give brainliest
Nikolay [14]
Answer: 90

Explanation:
7 0
3 years ago
Read 2 more answers
For 0 ≤ θ &lt; 2 π what are the solutions to sin^2(θ) =2sin^2(θ/2)
gregori [183]

Recall the half-angle identity for sine,

sin²(x/2) = (1 - cos(x))/2

Then the given equation is identical to

sin²(θ) = 1 - cos(θ)

Also recall the Pythagorean identity,

sin²(θ) + cos²(θ) = 1

Then we rewrite the equation as

1 - cos²(θ) = 1 - cos(θ)

Factoring the left side, we have

(1 - cos(θ)) (1 + cos(θ)) = 1 - cos(θ)

and so

(1 - cos(θ)) (1 + cos(θ)) - (1 - cos(θ)) = 0

and we factor this further as

(1 - cos(θ)) (1 + cos(θ) - 1) = 0

which gives

cos(θ) (1 - cos(θ)) = 0

Then either

cos(θ) = 0   or   1 - cos(θ) = 0

cos(θ) = 0   or   cos(θ) = 1

[θ = arccos(0) + 2nπ   or   θ = -arccos(0) + 2nπ]

…   or   [θ = arccos(1) + 2nπ   or   θ = -arccos(1) + 2nπ]

(where n is any integer)

[θ = π/2 + 2nπ   or   θ = -π/2 + 2nπ]   or   [θ = 0 + 2nπ]

In the interval 0 ≤ θ < 2π, we get three solutions:

• first solution set with n = 0   ⇒   θ = π/2

• second solution set with n = 1   ⇒   θ = 3π/2

• third solution set with n = 0   ⇒   θ = 0

So, the first choice is correct.

6 0
3 years ago
What is the place value of 3 of 652.4397. N Place Value: Select an answer Select an answer Get help: Vi tenths Points possible h
Stells [14]

Answer:

100 thousandths

Step-by-step explanation:

7 0
2 years ago
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