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Gnesinka [82]
3 years ago
7

Please answer both questions and show all work.

Mathematics
1 answer:
sammy [17]3 years ago
3 0
(1) 6 1/3 = 76/12
2 3/4 =  33/12
___________+
109/12
divide and get 9 1/12
-----------------------------------
(2)3 1/2 = 43/12
5 1/2 = 66/12
*from first answer* 109/12
add and you get 218/12
divide and get the answer 18 1/6
Hope this helps you!
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In which quadrant is (, -1.8)?
xenn [34]
Should be 2nd quadrant
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3 years ago
If y varies directly with x when y=1/3 and x = 12, find x if y is 4.
Nikitich [7]

Answer:

x = 144

Step-by-step explanation:

y α x

we introduce a constant 'k' when changing from α to =.

i.e y = kx

y = ⅓ when x = 12

Meaning, ⅓ = k * 12

⅓ = 12k

k = ⅓ ÷ 12

k = 1/36

If y = 4, x = ?

using the equation y = kx

4 = (1/36)x

x = 4 ÷ (1/36)

x = 4 * 36

x = 144

6 0
3 years ago
Pls answer what you can and I will give brainliest.
belka [17]

Answer:

I believe the top one is true! I'm sorry I couldn't answer them all :c

Step-by-step explanation:

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2 years ago
What is the scientific notation of each number 6908100000
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Answer:

6.9081 x 10^{9}

Step-by-step explanation:

7 0
2 years ago
find two positive real numbers such that they sum to 108 and the product of the first times the square of the second is a maximu
topjm [15]

The two positive numbers are 36 and 72 which gives a sum equal to 108 and the product of 36 and the square of 72 is a maximum.

Any integer greater than zero is considered a positive number. A positive number can either be written as a number or with the "+" symbol in front of it.

Let us consider the two positive real numbers as x and y. Then, their sum is written as,

x+y=108

Then, y=108-x

And the product is written as,

P=xy²

Substitute value of y in the above equation, we get,

\begin{aligned}P&=x(108-x)^2\\&=x(11664-216x+x^2)\\&=11664x-216x^2+x^3\\&=x^3-216x^2+11664x\end{aligned}

Now, differentiate P with respect to x, and we get,

\begin{aligned}\frac{dP}{dx}&=3x^2-432x+11664\\&=x^2-144x+3888\end{aligned}

Solving the above equation to zero, we get,

\begin{aligned}x^2-144x+3888&=0\\x^2-36x-108x+3888&=0\\x(x-36)-108(x-36)&=0\\(x-108)(x-36)&=0\\x&=\text{108 or 36}\end{aligned}

Substitute values of x in y=108-x, to get values of y.

If we substitute x=108, we get the y value as zero which doesn't give the required solution.

But, if we substitute x=36, we get,

\begin{aligned}y&=108-36\\y&=72\end{aligned}

Thus, the two positive numbers are 36 and 72.

To know more about positive numbers:

brainly.com/question/1635103

#SPJ4

4 0
11 months ago
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