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4 years ago

5

A ball is thrown and the height in feet of the ball is given by 1

1 answer:

MakcuM [25]4 years ago

7 0

H = -16t^2 + 40t + 11

we need to covert this to vertex form

h = -16(t^2 - 2.5t) + 11

= -16 [(t - 1.25)^2 - 1.5625] + 11

= -16(t-1.25)^2 + 25 + 11

= -16(t - 1.25)^2 + 36

The answer is 36 feet

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Let A = {a, b, c}, B = {b, c, d}, and C = {b, c, e}. (a) Find A ∪ (B ∩ C), (A ∪ B) ∩ C, and (A ∪ B) ∩ (A ∪ C). (Enter your answe

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(a)

$A\ u\ (B\ n\ C) = \{a,b,c\}$

$(A\ u\ B)\ n\ C = \{b,c\}$

$(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}$

$(A\ u\ B)\ n\ C = (A\ u\ B)\ n\ (A\ u\ C)$

(b)

$A\ n\ (B\ u\ C) = \{b,c\}$

$(A\ n\ B)\ u\ C = \{b,c,e\}$

$(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}$

$A\ n\ (B\ u\ C) = (A\ n\ B)\ u\ (A\ n\ C)$

(c)

$(A - B) - C = \{a\}$

$A - (B - C) = \{a,b,c\}$

<em>They are not equal</em>

<em></em>

Step-by-step explanation:

Given

$A= \{a,b,c\}$

$B =\{b,c,d\}$

$C = \{b,c,e\}$

Solving (a):

$A\ u\ (B\ n\ C)$

$(A\ u\ B)\ n\ C$

$(A\ u\ B)\ n\ (A\ u\ C)$

$A\ u\ (B\ n\ C)$

B n C means common elements between B and C;

So:

$B\ n\ C = \{b,c,d\}\ n\ \{b,c,e\}$

$B\ n\ C = \{b,c\}$

So:

$A\ u\ (B\ n\ C) = \{a,b,c\}\ u\ \{b,c\}$

u means union (without repetition)

So:

$A\ u\ (B\ n\ C) = \{a,b,c\}$

Using the illustrations of u and n, we have:

$(A\ u\ B)\ n\ C$

$(A\ u\ B)\ n\ C = (\{a,b,c\}\ u\ \{b,c,d\})\ n\ C$

Solve the bracket

$(A\ u\ B)\ n\ C = (\{a,b,c,d\})\ n\ C$

Substitute the value of set C

$(A\ u\ B)\ n\ C = \{a,b,c,d\}\ n\ \{b,c,e\}$

Apply intersection rule

$(A\ u\ B)\ n\ C = \{b,c\}$

$(A\ u\ B)\ n\ (A\ u\ C)$

In above:

$A\ u\ B = \{a,b,c,d\}$

Solving A u C, we have:

$A\ u\ C = \{a,b,c\}\ u\ \{b,c,e\}$

Apply union rule

$A\ u\ C = \{b,c\}$

So:

$(A\ u\ B)\ n\ (A\ u\ C) = \{a,b,c,d\}\ n\ \{b,c\}$

$(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}$

<u>The equal sets</u>

We have:

$A\ u\ (B\ n\ C) = \{a,b,c\}$

$(A\ u\ B)\ n\ C = \{b,c\}$

$(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}$

So, the equal sets are:

$(A\ u\ B)\ n\ C$ and $(A\ u\ B)\ n\ (A\ u\ C)$

They both equal to $\{b,c\}$

So:

$(A\ u\ B)\ n\ C = (A\ u\ B)\ n\ (A\ u\ C)$

Solving (b):

$A\ n\ (B\ u\ C)$

$(A\ n\ B)\ u\ C$

$(A\ n\ B)\ u\ (A\ n\ C)$

So, we have:

$A\ n\ (B\ u\ C) = \{a,b,c\}\ n\ (\{b,c,d\}\ u\ \{b,c,e\})$

Solve the bracket

$A\ n\ (B\ u\ C) = \{a,b,c\}\ n\ (\{b,c,d,e\})$

Apply intersection rule

$A\ n\ (B\ u\ C) = \{b,c\}$

$(A\ n\ B)\ u\ C = (\{a,b,c\}\ n\ \{b,c,d\})\ u\ \{b,c,e\}$

Solve the bracket

$(A\ n\ B)\ u\ C = \{b,c\}\ u\ \{b,c,e\}$

Apply union rule

$(A\ n\ B)\ u\ C = \{b,c,e\}$

$(A\ n\ B)\ u\ (A\ n\ C) = (\{a,b,c\}\ n\ \{b,c,d\})\ u\ (\{a,b,c\}\ n\ \{b,c,e\})$

Solve each bracket

$(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}\ u\ \{b,c\}$

Apply union rule

$(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}$

<u>The equal set</u>

We have:

$A\ n\ (B\ u\ C) = \{b,c\}$

$(A\ n\ B)\ u\ C = \{b,c,e\}$

$(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}$

So, the equal sets are:

$A\ n\ (B\ u\ C)$ and $(A\ n\ B)\ u\ (A\ n\ C)$

They both equal to $\{b,c\}$

So:

$A\ n\ (B\ u\ C) = (A\ n\ B)\ u\ (A\ n\ C)$

Solving (c):

$(A - B) - C$

$A - (B - C)$

This illustrates difference.

$A - B$ returns the elements in A and not B

Using that illustration, we have:

$(A - B) - C = (\{a,b,c\} - \{b,c,d\}) - \{b,c,e\}$

Solve the bracket

$(A - B) - C = \{a\} - \{b,c,e\}$

$(A - B) - C = \{a\}$

Similarly:

$A - (B - C) = \{a,b,c\} - (\{b,c,d\} - \{b,c,e\})$

$A - (B - C) = \{a,b,c\} - \{d\}$

$A - (B - C) = \{a,b,c\}$

<em>They are not equal</em>

4 0

3 years ago