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Bad White [126]

3 years ago

9

julio bought 3/5 pound of potto salad for dinner wilma bought 5/4 the amount of potato salad that julio ei how much potato salad

did wilma

1 answer:

Harrizon [31]3 years ago

6 0

Wilma brought 3/4 pounds of potato salad. When you multiply the fractions, you get 15/20, which can simplify to three fourths

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EXAMPLE problem: 9.1 divided by 3576 rounded to the nearest whole number

Svetradugi [14.3K]

We have to divide 9.1 by 3576:
9.1 : 3576 = 0.0025447427
And rounded to the nearest whole number:
0.0025447427 ≈ 0

8 0

3 years ago

Read 2 more answers

Find the slope of DE and FG

Pavlova-9 [17]

Answer:

Part 1) The slope of DE is $m=\frac{c}{a+b}$

Part 2) The slope of FG is $m=\frac{c}{a+b}$

Step-by-step explanation:

we know that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

step 1

Find the slope of DE

we have

D(-a-b,c) and E(0,2c)

substitute in the formula

$m=\frac{2c-c}{0-(-a-b)}$

$m=\frac{c}{a+b}$

step 2

Find the slope of FG

we have

F(a+b,c) and G(0,0)

substitute in the formula

$m=\frac{0-c}{0-(a+b)}$

$m=\frac{-c}{-(a+b)}$

Simplify the sign minus

$m=\frac{c}{a+b}$

Note The slope of DE is equal to the slope of FG. therefore DE and FG are parallel

7 0

3 years ago

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

Which lines or segments are parallel based off markings?

liberstina [14]

AD BC are parallel because the lines are equal

7 0

3 years ago

The table shows five transactions and the resulting account balance in a bank account, except some numbers are missing. Fill the

Allushta [10]

Transaction 3= -155.15
Transaction 4= -223.75

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3 years ago