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Unions and intersections of sets

f(x)

x-1

, the remainder is

12

f(x)

x-4

, the remainder is

3

. Find the remainder when

f(x)

x^{2}-5 x+4

\newline

A=1,2,4,8,9

\newline

B=8,9,10,12

\newline

A \cap B

\newline

\{1,2,4,8,9,10,12\}

\newline

\{8,9\}

\newline

\{1,2,4,8,10\}

\newline

\{1,2,4\}

1

a=3

b=5

2 a+b

\newline

10

\newline

16

\newline

13

\newline

11

\newline

6

: The product of

82

5

\newline

400

\newline

410

\newline

420

D = \{-6, -2, 1, 5, 6, 11\}

E = \{-6, -2, 1, 4, 5, 11\}

D \cap E

\newline

\newline

\{-6, -2, 1, 5, 11\}

\newline

\{-6, -2, 1, 4, 5, 6, 11\}

\newline

\{-6, 1, 4, 5\}

\newline

\{-6, 5\}

G = \{-3, -1, 0, 5\}

H = \{0, 3, 5\}

G \cap H

\newline

\newline

\{0, 5\}

\newline

\{-3, -1, 0, 3, 5\}

\newline

\{5\}

\newline

\{0, 3, 5\}

M = \{2, 4, 6, 8, 10\}

N = \{2, 6, 8, 10, 14, 16\}

M \cap N

\newline

\newline

\{2, 6, 8, 10, 14, 16\}

\newline

\{6, 8, 10\}

\newline

\{2, 4, 6, 8, 10, 14, 16\}

\newline

\{2, 6, 8, 10\}

S = \{1, 3, 5, 7, 9\}

T = \{5, 8, 9\}

S \cap T

\newline

\newline

\{5, 9\}

\newline

\{5\}

\newline

\{1, 3, 5, 7, 8, 9\}

\newline

\{3, 5, 8\}

R = \{7\}

S = \{2, 3, 4, 5, 6\}

R \cup S

\newline

\newline

\{2, 3, 5, 6\}

\newline

\{2, 7\}

\newline

\emptyset

\newline

\{2, 3, 4, 5, 6, 7\}

S = \{3\}

T = \{2, 8, 9, 14\}

S \cup T

\newline

\newline

\{2, 3, 8, 9, 14\}

\newline

\{3, 8, 9, 14\}

\newline

\{2, 3, 9\}

\newline

\emptyset

S = \{3\}

T = \{2, 8, 9, 14\}

S \cup T

\newline

\newline

\{3, 8, 9, 14\}

\newline

\emptyset

\newline

\{2, 3, 8, 9, 14\}

\newline

\{2, 3, 9\}

A = \{0\}

B = \{-5, -2, 0, 9\}

A \cup B

\newline

\newline

\{0\}

\newline

\{0, 9\}

\newline

\{-5, -2, 9\}

\newline

\{-5, -2, 0, 9\}

P = \{-1\}

Q = \{-7, -5, -3, -1, 1\}

P \cup Q

\newline

\newline

\{-1\}

\newline

\{-5, -3\}

\newline

\{-7, -5, -3, -1, 1\}

\newline

\{-7, -5, -3, 1\}

M = \{2, 4, 6, 8, 10\}

N = \{2, 6, 8, 10, 14, 16\}

M \cap N

\newline

\newline

\{6, 8, 10\}

\newline

\{2, 6, 8, 10, 14, 16\}

\newline

\{2, 6, 8, 10\}

\newline

\{2, 4, 6, 8, 10, 14, 16\}

S = \{3\}

T = \{2, 8, 9, 14\}

S \cup T

\newline

\newline

\{3, 8, 9, 14\}

\newline

\{2, 3, 9\}

\newline

\{2, 3, 8, 9, 14\}

\newline

\emptyset

D = \{0, 1, 5, 7, 8\}

E = \{1, 8\}

D \cup E

\newline

\newline

\{0, 1, 5, 7, 8\}

\newline

\{1\}

\newline

\{1, 8\}

\newline

\{0, 5, 7\}

M = \{2, 4, 6, 8, 10\}

N = \{2, 6, 8, 10, 14, 16\}

M \cap N

\newline

\newline

\{2, 6, 8, 10\}

\newline

\{2, 4, 6, 8, 10, 14, 16\}

\newline

\{6, 8, 10\}

\newline

\{2, 6, 8, 10, 14, 16\}

S = \{1, 3, 5, 7, 9\}

T = \{5, 8, 9\}

S \cap T

\newline

\newline

\{1, 3, 5, 7, 8, 9\}

\newline

\{5, 9\}

\newline

\{5\}

\newline

\{3, 5, 8\}

D = \{0, 1, 5, 7, 8\}

E = \{1, 8\}

D \cup E

\newline

\newline

\{0, 5, 7\}

\newline

\{1\}

\newline

\{1, 8\}

\newline

\{0, 1, 5, 7, 8\}

D = \{0, 1, 5, 7, 8\}

E = \{1, 8\}

D \cup E

\newline

\newline

\{0, 1, 5, 7, 8\}

\newline

\{0, 5, 7\}

\newline

\{1, 8\}

\newline

\{1\}

\begin{array}{l} 8 y=4 x^{2}-12 x+46 \\ y=\frac{3}{2} x+\frac{5}{4} \end{array}

\newline

(a, b)

is the solution to the system of equations shown, what is the value of

a

8p +3q-18

p=\frac{1}{2}

q=7

7j + 5 - 8k

j= 0.5

k= 0.25

10h+30+50j

to identify the equivalent expressions.

\newline

2

\newline

5(2h+6+10j)

\newline

10(h+3+5j)

\newline

2(5h+15+25j)

\newline

15(\frac{2}{3}h+2+\frac{10}{3}j)

\mathrm{E}=\left[\begin{array}{lll}5 & 5 & 1 \\ 4 & 0 & 3\end{array}\right]

\mathrm{B}=\left[\begin{array}{rr}-1 & 5 \\ 1 & 2 \\ -2 & 3\end{array}\right]

\newline

\mathrm{H}=\mathrm{EB}

\mathrm{H}

\newline

\mathbf{H}=

\mathrm{D}=\left[\begin{array}{rrr}-1 & 0 & 4 \\ 2 & 0 & 0\end{array}\right]

F=\left[\begin{array}{rr}-1 & 1 \\ -1 & 3 \\ 2 & 4\end{array}\right]

\newline

\mathrm{H}=\mathrm{DF}

\mathrm{H}

\newline

\mathbf{H}=

\mathrm{E}=\left[\begin{array}{rrr}1 & -1 & 1 \\ 2 & 3 & 4\end{array}\right]

A=\left[\begin{array}{rr}2 & 4 \\ 2 & -2 \\ -1 & 5\end{array}\right]

\newline

\mathrm{H}=\mathrm{EA}

\mathrm{H}

\newline

\mathbf{H}=

A=\left[\begin{array}{r} -1 \\ 4 \\ 4 \end{array}\right] \text { and } \mathrm{F}=\left[\begin{array}{ll} 0 & -2 \end{array}\right]

\newline

\mathrm{H}=\mathrm{AF}

\mathrm{H}

\newline

\mathbf{H}=

E=\left[\begin{array}{lll}1 & 4 & 2 \\ 1 & 2 & 2\end{array}\right]

\mathrm{B}=\left[\begin{array}{rr}1 & 3 \\ 2 & 3 \\ 2 & -2\end{array}\right]

\newline

\mathrm{H}=\mathrm{EB}

\mathrm{H}

\newline

\mathbf{H}=

\mathrm{E}=\left[\begin{array}{rr}-1 & 4 \\ 5 & 2\end{array}\right]

C=\left[\begin{array}{rr}1 & 2 \\ -1 & 1\end{array}\right]

\newline

\mathrm{H}=\mathrm{EC}

\mathrm{H}

\newline

\mathbf{H}=

\mathrm{B}=\left[\begin{array}{ll} 1 & 4 \\ 1 & 3 \end{array}\right] \text { and } \mathrm{D}=\left[\begin{array}{rr} 4 & 3 \\ 0 & -1 \end{array}\right]

\newline

\mathrm{H}=\mathrm{BD}

\mathrm{H}

\newline

\mathbf{H}=

\mathrm{D}=\left[\begin{array}{rr}1 & 0 \\ -1 & 5\end{array}\right]

\mathrm{B}=\left[\begin{array}{lll} 1 & 0 & 2 \\ 4 & 5 & 4 \end{array}\right]

\newline

\mathrm{H}=\mathrm{DB}

\mathrm{H}

\newline

\mathbf{H}=

\mathrm{F}=\left[\begin{array}{ll} 0 & 4 \\ 5 & 2 \end{array}\right] \text { and } \mathrm{D}=\left[\begin{array}{ll} 2 & 5 \\ 3 & 0 \end{array}\right]

\newline

\mathrm{H}=\mathrm{FD}

\mathrm{H}

\newline

\mathbf{H}=

\mathrm{F}=\left[\begin{array}{l} 5 \\ 0 \\ 4 \end{array}\right] \text { and } \mathrm{A}=\left[\begin{array}{ll} -1 & 1 \end{array}\right]

\newline

\mathrm{H}=\mathrm{FA}

\mathrm{H}

\newline

\mathbf{H}=

\mathrm{E}=\left[\begin{array}{r} 5 \\ 2 \\ -2 \end{array}\right] \text { and } \mathrm{D}=\left[\begin{array}{ll} 5 & 3 \end{array}\right]

\newline

\mathrm{H}=\mathrm{ED}

\mathrm{H}

\newline

\mathbf{H}=

\mathrm{B}=\left[\begin{array}{r}0 \\ -1 \\ 2\end{array}\right]

\mathrm{A}=\left[\begin{array}{ll}1 & 2\end{array}\right]

\newline

\mathrm{H}=\mathrm{BA}

\mathrm{H}

\newline

\mathbf{H}=

\mathbf{B}=\left[\begin{array}{rr}5 & -2 \\ -1 & 5\end{array}\right]

\mathrm{A}=\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]

\newline

\mathrm{H}=\mathrm{BA}

\mathrm{H}

\newline

\mathbf{H}=

E=\left[\begin{array}{ll}-1 & -2 \\ -2 & -1\end{array}\right]

A=\left[\begin{array}{rr} 2 & 0 \\ 2 & -1 \end{array}\right] .

\newline

\mathrm{H}=\mathrm{EA}

\mathrm{H}

\newline

\mathbf{H}=

-a+(-b)

a=6.05

b=3.611

f-g+(-2)

f=-3.005

g=4.7

Consider the expression

4(8 x+5)

\newline

2

descriptions of the parts of the expression.

\newline

1

. The entire expression is the product of

\square

\newline

2

(8 x+5)

\square

x^{2}+6 x-5=0

, what are the values of

x

\newline

1

\newline

-3-\sqrt{14}

-3+\sqrt{14}

\newline

3-\sqrt{14}

3+\sqrt{14}

\newline

-1

-5

\newline

1

5

What is the product of

6, x^{2}-3 x

x-10

\newline

1

\newline

-42 x+180

\newline

6 x^{2}-78 x+180

\newline

6 x^{3}-42 x^{2}+180 x

\newline

6 x^{3}-78 x^{2}+180 x

x+y=-4

x-y=2

, what is the value of

(x+y)\left(x^{2}-y^{2}\right) ?

f(x)=3 x-1

g(x)=x^{2}+1

, what is the value of

g(f(3))

\newline

1

\newline

8

\newline

10

\newline

29

\newline

65

\begin{array}{c} 3-m=2(\ell-4) \\ m=\ell-4 \end{array}

\newline

Consider the given system of equations. If

(\ell, m)

is the solution to the system, then what is the value of

\ell \cdot m

2+2(8347-4783)=-2 j

, then what is the value of

-j

What is the value of

y^{2}-x^{4}

y=8

x=2

What is the value of

y^{2}-x^{2}

x=5

y=6

What is the value of

x^{6}+y^{2}

x=1

y=9