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Anestetic [448]

2 years ago

10

Suppose that you have a bowl of 500 m&m candies, and each day you eat 1/4 of the candies you have. Is the number of candies

left changing linearly or exponentially? Write an equation to model the number of candies left after n days.

1 answer:

Aleks [24]2 years ago

8 0

Answer:

$500(.75)^{n}$

Step-by-step explanation:

It is changing exponentially because if you are eating a fraction of the left over m & m's, that is a different amount each time so it can't be linear.

500(1-(1/4))^n

$500(.75)^{n}$

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A kangaroo hops 2 kilometers in 4 minutes. How far can the Kangaroo hop in 10 minutes?

Artyom0805 [142]

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Step-by-step explanation:

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3 years ago

Read 2 more answers

A) Work out the value of 5^2 – 1^5× 2^4<br>b) Work out the value of 7^2 + 4^3 ÷ 2^5

NemiM [27]

First you have to know BODMAS/BIDMAS (idk if all schools teach it like this) but this is the order in which you complete questions so it’s Brackets, Indices, Division/Multiplication, Addition/Subtraction
a) 5^2 is 5x5 which equals 25 1 to the power of anything is always 1 and 2^4 is 2x2x2x2 which is 16
So 5^2-1^5x2^4 = 25-1x16 which is then 25-16 = 9
b) 7^2+4^3 divided by 2^5 so 7^2 is 7x7 which = 49 then 4^3 means 4x4x4 = 64 and 2^5 is 2x2x2x2x2 which equals 32 so
7^2+4^3 divides by 2^5 is 49+64/32 = 49+2 which is 51

8 0

3 years ago

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pshichka [43]

Answer:

325.26

Step-by-step explanation:

that's the answer because you have to multiply the top

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3 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago