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BabaBlast [244]
2 years ago
7

Contienen informacion sobre la fertilidad ,resistencia de medicamentos dañinos para la celula

Mathematics
1 answer:
Vladimir79 [104]2 years ago
6 0

Answer:

niwim nutkiwmroi jtfrekm

Step-by-step explanation:

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Factor the following expression. 36y - 12
Over [174]

Hey there!


36y - 12


FIRST LOOK AT THE GCF (Greatest Common Factor)


36y: 1, 2, 3, 4, 6, 9, 12, 18, 36, and y


12: 1, 2, 3, 4, 6, and 12


GCF: 12


So, 12 would most likely be number before you make your parentheses


12(?y - 1)


The numbers in your original equation both share 3 & -1


So, therefore the factor form of your equation should be: 12(3y - 1)


Answer: 12(3y - 1)


Good luck on your assignment & enjoy your day!


~Amphitrite1040:)

7 0
2 years ago
Read 2 more answers
Each equation below is followed by several stories.
zloy xaker [14]
Not sure sorry edit : oh sorry I was suppose to add this in the comment section
7 0
3 years ago
YOU MUST FOLLOW THE WORD PROBLEM 5 STEP PROCESS
madreJ [45]

Answer:

$10,100 000,000.006 but it says round to the dollar so it is $10,100 000,000

Step-by-step explanation:

136/100 × $7,426,470,588.24

8 0
2 years ago
I’ll mark you brainliest!!!
notka56 [123]
Answer:
C. (1,18)
Explanation
If you don’t know how to find it, just plug in the values into the equation until they are equal to each other
3 0
3 years ago
Read 2 more answers
Use Newton’s Method to find the solution to x^3+1=2x+3 use x_1=2 and find x_4 accurate to six decimal places. Hint use x^3-2x-2=
luda_lava [24]

Let f(x) = x^3 - 2x - 2. Then differentiating, we get

f'(x) = 3x^2 - 2

We approximate f(x) at x_1=2 with the tangent line,

f(x) \approx f(x_1) + f'(x_1) (x - x_1) = 10x - 18

The x-intercept for this approximation will be our next approximation for the root,

10x - 18 = 0 \implies x_2 = \dfrac95

Repeat this process. Approximate f(x) at x_2 = \frac95.

f(x) \approx f(x_2) + f'(x_2) (x-x_2) = \dfrac{193}{25}x - \dfrac{1708}{125}

Then

\dfrac{193}{25}x - \dfrac{1708}{125} = 0 \implies x_3 = \dfrac{1708}{965}

Once more. Approximate f(x) at x_3.

f(x) \approx f(x_3) + f'(x_3) (x - x_3) = \dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125}

Then

\dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125} = 0 \\\\ \implies x_4 = \dfrac{5,881,319,037}{3,324,107,515} \approx 1.769292663 \approx \boxed{1.769293}

Compare this to the actual root of f(x), which is approximately <u>1.76929</u>2354, matching up to the first 5 digits after the decimal place.

4 0
2 years ago
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