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2 years ago

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The zoo needed to raise $25,000. Zoo officials planned a dinner/dance event and charged $300 per couple. What is the fewest numb

er of couples that would need to attend in order for the goal to be met

1 answer:

Zielflug [23.3K]2 years ago

3 0

84 the least hope that helps

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30 fat points. Rita builds a pool in her backyard. The pool measures 60 feet long, 32 feet wide and 8 feet deep. How much water

Brums [2.3K]

Answer:

7 quarts

Step-by-step explanation:

8 0

2 years ago

What substitution should be used to rewrite 6(x + 5)2 + 5(x + 5) – 4 = 0 as a quadratic equation? u = (x + 5) u = (x – 5) u = (x

Pachacha [2.7K]

6(x+5)² + 5(x+5)-4 = 0

(x+5)(x+5) = x(x+5)+5(x+5) = x² + 5x + 5x + 25 = x² + 10x + 25

6(x²+10x+25) + 5(x+5) - 4 = 0
6x² + 60x + 150 + 5x + 25 - 4 = 0
6x² + 60x + 5x + 150 + 25 - 4 = 0
6x² + 65x + 171 = 0

<span>u = (x + 5)</span>

5 0

2 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

1 year ago

Zachary is buying 4 tires for his car. The table shows the prices and the advertised sales for the same type of tire at 4 tire s

balandron [24]

Answer:

prolly r

Step-by-step explanation:

I guessed bc this looks really hard n borin

0 0

2 years ago

contractor wishes to build 9 houses, each different in design. In how many ways can he place these houses on a street if 6 lots

Tju [1.3M]

The houses can be placed in 362,880 ways.

<u>Step-by-step explanation:</u>

The 9 houses are each in different design.

The each lot can place any of the 9 houses.

The 1st lot can place anyone house of all the 9 houses.
The 2nd lot can place one of remaining 8 houses.
The 3rd lot can place one of remaining 7 houses.

Similarly, the process gets repeated until the last house is placed on a lot.

<u>From the above steps, it can be determined that :</u>

The number of ways to place the 9 houses in 9 lots = 9!

⇒ 9×8×7×6×5×4×3×2×1

⇒ 362880 ways.

Therefore, the houses can be placed in 362880 ways.

5 0

2 years ago