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erastovalidia [21]
3 years ago
5

if each bag holds 3.75 pounds and they use 0.75 pound of birdseed each each day, for how many will one bag last

Mathematics
1 answer:
aniked [119]3 years ago
4 0
It will last 5 days. That's what I got.  :D
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The vertex of this parabola is at (-3, -1). When the y-value is 0, the x-value is 4. What is the coefficient of the squared expr
Klio2033 [76]

Answer:

S(-3, -1)

P(4, 0)

a = (Py - Sy) / (Px - Sx)^2

a = (0 - (-1)) / (4 - (-3))^2

a = 1 / 7^2

<u>a = 1 / 49</u>

6 0
3 years ago
Read 2 more answers
A radioactive substance decays exponentially. A scientist begins with 110 milligrams of a radioactive substance. After 20 hours,
SpyIntel [72]

Answer:

  44.7 mg

Step-by-step explanation:

The equation for exponential decay can be written in the form ...

  y = a·b^(t/p)

where 'a' is the initial value, 'b' is the decay factor, 'p' is the period over which the decay factor is applicable, and t is time in the same units as p.

<h3>Setup</h3>

Using the above equation, we have ...

  a = initial value = 110 mg

  b = decay factor = 55/110 = 1/2 over time period p=20 hours

Then the equation is ...

  y = 110·(1/2)^(t/20) . . . . amount remaining after t hours

<h3>Solution</h3>

We want the amount remaining after 26 hours. That will be ...

  y = 110·(1/2)^(26/20) ≈ 44.67

About 44.7 milligrams will remain after 26 hours.

4 0
2 years ago
X^2-8x-36=0<br> solve by completing the square <br> ANSWER ASAP
larisa86 [58]

Answer:

hope it helps

7 0
3 years ago
PLEASE HELP ASAP.
Dafna11 [192]
<span>c. JK

that is where the two planes intersect at

I hope this helps, good luck! :)</span>
6 0
4 years ago
Read 2 more answers
I do not know how to solve this problem or know what to do with the exponents
horsena [70]
K, remember
(ab)/(cd)=(a/c)(b/d) or whatever
also
(ab)^c=(a^c)(b^c)
and
x^{-m}= \frac{1}{x^m}
and
x^ \frac{m}{n}= \sqrt[n]{x^m}
and
( \frac{x}{y} )^m= \frac{x^m}{y^m}
and
(x^m)^n=x^{mn}
and
(a/b)/(c/d)=(a/b)(d/c)=(ad)/(bc)


so
( \frac{-7x^ \frac{3}{2} }{5y^4} )^{-2}=
( \frac{-7}{5} )^{-2}( \frac{x^ \frac{3}{2} }{y^4} )^{-2}=
( \frac{(-7)^{-2}}{5^{-2}} )( \frac{(x^ \frac{3}{2})^{-2} }{(y^4)^{-2}} )=
( \frac{ \frac{1}{(-7)^2} }{ \frac{1}{5^2} } )( \frac{x^ \frac{-6}{2} }{y^{-8}} )=
( \frac{ \frac{1}{49} }{ \frac{1}{25} } )( \frac{(x^{-3}) }{\frac{1}{y^8}} )
( \frac{ \frac{1}{49} }{ \frac{1}{25} } )( \frac{\frac{1}{x^3} }{\frac{1}{y^8}} )=
(\frac{25}{49} )( \frac{y^8}{x^3}=
\frac{25y^8}{49x^3}
4 0
3 years ago
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