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natulia [17]
3 years ago
14

What is 3x-2y=4 when written in slope intercept form

Mathematics
1 answer:
Verdich [7]3 years ago
7 0

Answer: y = 3/2x - 2

Step-by-step explanation:

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Please someone help me! I know that the answer is close to 17, so if u know the answer please show me how to do it!
gogolik [260]

Answer:

Ask ur mom she's nice

Step-by-step explanation:


8 0
3 years ago
I will mark you as brainliest if you're answer is correct​
GalinKa [24]

Answer:

perimeter = 28.13 cm

Step-by-step explanation:

Calculate the length of a line drawn from A to C.

sin 85 = b/8, b = 7.97

cos 85 = c/8, c = 0.697

AC² =7.97² + (6 + 0.697)² = 63.52 + 44.85 = 108.37

AC = 10.41

Now you have a right triangle ADC with a known hypotenuse and  one leg.

Using the Pythagorean theorem:

DC² = 10.41² - 5²

DC² = 108.37 - 25 = 83.37

DC = 9.13 cm

perimeter = 5 + 6 + 8 + 9.13 = 28.13 cm

8 0
3 years ago
The table below represents the atmospheric temperature at a location as a function of the altitude:
RSB [31]

from x15 to x 25 is 4 - -16 = -20 degree change

25 -15 = 10000 feet

-20/10 = -2 degrees every 1000 feet


7 0
4 years ago
Does anybody know how to do time with exponential decay
My name is Ann [436]
<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

b_n=0.12\times b_{n-1}

Why does this work? Initially, you start with 500 mL of air that you breathe in, so b_1=500\text{ mL}. After the second breath, you have 12% of the original air left in your lungs, or b_2=0.12\timesb_1=0.12\times500=60\text{ mL}. After the third breath, you have b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}, and so on.

You can find the amount of original air left in your lungs after n breaths by solving for b_n explicitly. This isn't too hard:

b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots

and so on. The pattern is such that you arrive at

b_n=0.12^{n-1}b_1

and so the amount of air remaining after 50 breaths is

b_{50}=0.12^{50-1}b_1=0.12^{49}\times500\approx3.7918\times10^{-43}

which is a very small number close to zero.</span>
5 0
4 years ago
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