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aniked [119]
3 years ago
11

A farmer feeds a cow 9300 milligrams of an antibiotic. Every hour, 50% of the drug breaks down in the cow's body. How much will

be left in 6 hours?
Mathematics
1 answer:
hichkok12 [17]3 years ago
6 0

9514 1404 393

Answer:

  about 145.3 mg

Step-by-step explanation:

If half breaks down in the hour, then half remains. That is, each hour the amount remaining is multiplied by 1/2. The remainder after 6 hours is ...

  (9300 mg)(1/2)^6 = 145.3125 gm

About 145.3 mg will be left after 6 hours.

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Clare made $160 babysitting last summer. She put the money in a savings account that pays 3% interest per year. If clare doesn't
AVprozaik [17]

We have been given that Clare made $160 babysitting last summer. She put the money in a savings account that pays 3% interest per year. If Clare doesn't touch the money in her account, she can find the amount she'll have the next year by multiplying her current amount 1.03.

We are asked to write an expression for the amount of money Clare would have after 30 years if she never withdraws money from her account.

We will use exponential growth function to solve our given problem.

An exponential growth function is in form y=a(1+r)^x, where

y = Final value,

a = Initial value,

r = Growth rate in decimal form,

x = Time.

3\%=\frac{3}{100}=0.03

We can see that initial value is $160. Upon substituting our given values in above formula, we will get:

y=160(1+0.03)^x

y=160(1.03)^x

To find amount of money in Clare's account after 30 years, we need to substitute x=30 in our equation.

y=160(1.03)^{30}

Therefore, the expression 160(1.03)^{30} represents the amount of money that Clare would have after 30 years.

8 0
3 years ago
Describe the process of subtracting a negative integer from a positive integer
emmasim [6.3K]
Its basically just subtracting from one another

for example, 6-6 is 6

or -4-3= 7
5 0
3 years ago
In a certain sequence of numbers, each term after the first is found by doubling and then adding $3$ to the previous term. If th
eimsori [14]
Given a_7=125 you can find a_6:

a_6= \frac{125-3}{2} =61
 a_5= \frac{61-3}{2} =29 \\ a_4= \frac{29-3}{2}=13 \\ a_3= \frac{13-3}{2} =5 \\ a_2= \frac{5-3}{2} =1 \\ a_1= \frac{1-3}{2}=-1
Answer: -1
6 0
3 years ago
Find all solutions to
BARSIC [14]

Answer:

x= 0 , \frac{1}{14} , \frac{-1}{12}

Step-by-step explanation:

Given, equation is \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x}. →→→ (1)

Now, by cubing the equation on both sides, we get

( \sqrt[3]{15x-1} + \sqrt[3]{13x+1} )³ = (4\sqrt[3]{x})³

⇒ (15x-1) + (13x+1) + 3×\sqrt[3]{15x-1}×\sqrt[3]{13x+1} (\sqrt[3]{15x-1} + \sqrt[3]{13x+1}) = 64 x.

⇒ 28x + 3×\sqrt[3]{15x-1}×\sqrt[3]{13x+1} (4\sqrt[3]{x}) = 64x.        

(since from (1),  \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x})

⇒ 12× \sqrt[3]{15x-1}×\sqrt[3]{13x+1} (\sqrt[3]{x})= 36x.

⇒ 3x = \sqrt[3]{(15x-1)(13+1)(x)}.

Now, once again cubing on both sides, we get

(3x)³ = (\sqrt[3]{(15x-1)(13+1)(x)})³.

⇒ 27x³ = (15x-1)(13x+1)(x).

⇒ 27x³ = 195x³ + 2x² - x

⇒ 168x³ + 2x² - x = 0

⇒ x(168x² + 2x -1) = 0

⇒ by, solving the equation we get ,

x = 0 ; x = \frac{1}{14} ; x = \frac{-1}{12}

therefore, solution is x= 0 , \frac{1}{14} , \frac{-1}{12}

7 0
3 years ago
Need help with this asap, please
Lilit [14]

Answer:

<h3>             Perimeter = 3√5 + 6 + √73 + √58 + 5√2</h3>

Step-by-step explanation:

AB=\sqrt{(6-0)^2+(5-8)^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt5\\\\BC=|-1-5|=|-6|=6\\\\CD=\sqrt{(-2-6)^2+(-4-(-1))^2}=\sqrt{64+9}=\sqrt{73}\\\\DE=\sqrt{(-5-(-2))^2+(3-(-4))^2}=\sqrt{9+49}=\sqrt{58}\\\\EA=\sqrt{(0-(-5))^2+(8-3)^2}=\sqrt{25+25}=\sqrt{25\cdot2}=5\sqrt2

6 0
3 years ago
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