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Eddi Din [679]
3 years ago
9

You received 1⁄3 pound of candy from your grandmother, 1⁄2 pound of candy from your sister, but your best friend ate 1⁄5 pound o

f candy. How much candy do you have left?
Mathematics
2 answers:
OverLord2011 [107]3 years ago
3 0
Give them the same denominator.
19/30 is what you have left
Vikentia [17]3 years ago
3 0
You set the same denominator which would become

2/6 and 3/6

5/6 is how much chocolate you get

Now you set the same denominator again

50/60 and 12/60

subtraction would cause

38/60 is amount of candy left
simplify to 19/30
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Answer:

12

Step-by-step explanation:

6+6=12

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Use partial products to multiply 35 times 77
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Find a polynomial $f(x)$ of degree $5$ such that both of these properties hold: $\bullet$ $f(x)$ is divisible by $x^3$. $\bullet
frosja888 [35]

There seems to be one character missing. But I gather that <em>f(x)</em> needs to satisfy

• x^3 divides f(x)

• (x-1)^3 divides f(x)^2

I'll also assume <em>f(x)</em> is monic, meaning the coefficient of the leading term is 1, or

f(x) = x^5 + \cdots

Since x^3 divides f(x), and

f(x) = x^3 p(x)

where p(x) is degree-2, and we can write it as

f(x)=x^3 (x^2+ax+b)

Now, we have

f(x)^2 = \left(x^3p(x)\right)^2 = x^6 p(x)^2

so if (x - 1)^3 divides f(x)^2, then p(x) is degree-2, so p(x)^2 is degree-4, and we can write

p(x)^2 = (x-1)^3 q(x)

where q(x) is degree-1.

Expanding the left side gives

p(x)^2 = x^4 + 2ax^3 + (a^2+2b)x^2 + 2abx + b^2

and dividing by (x-1)^3 leaves no remainder. If we actually compute the quotient, we wind up with

\dfrac{p(x)^2}{(x-1)^3} = \underbrace{x + 2a + 3}_{q(x)} + \dfrac{(a^2+6a+2b+6)x^2 + (2ab-6a-8)x +2a+b^2+3}{(x-1)^3}

If the remainder is supposed to be zero, then

\begin{cases}a^2+6a+2b+6 = 0 \\ 2ab-6a-8 = 0 \\ 2a+b^2+3 = 0\end{cases}

Adding these equations together and grouping terms, we get

(a^2+2ab+b^2) + (2a+2b) + (6-8+3) = 0 \\\\ (a+b)^2 + 2(a+b) + 1 = 0 \\\\ (a+b+1)^2 = 0 \implies a+b = -1

Then b=-1-a, and you can solve for <em>a</em> and <em>b</em> by substituting this into any of the three equations above. For instance,

2a+(-1-a)^2 + 3 = 0 \\\\ a^2 + 4a + 4 = 0 \\\\ (a+2)^2 = 0 \implies a=-2 \implies b=1

So, we end up with

p(x) = x^2 - 2x + 1 \\\\ \implies f(x) = x^3 (x^2 - 2x + 1) = \boxed{x^5-2x^4+x^3}

6 0
2 years ago
Solve<br><br> 1/5 (x+3)=−(2x+3)
Rasek [7]

Answer:

Step-by-step explanation:

just simplify the LHS first.

You can either multiply 1/5 by (x+3) and then solve

or

multiply both sides by 5 to get rid of 1/5 on LHS

I will multiply by 5

(x+3)= -10x-15     ( 5*1/5(x+3)= -5(2x+3)

now rearrange the equation

x+3=-10x-15

-10x-15-x-3=0

-11x-17=0

-11x=17

x= -17/11

4 0
2 years ago
Find 32% of $115.00.
mel-nik [20]

Answer:

$36.8

Step-by-step explanation:

5 0
2 years ago
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