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VikaD [51]
3 years ago
8

Drag the tiles to the boxes to form correct pairs.

Mathematics
2 answers:
denis23 [38]3 years ago
3 0

Answer:

Answer is in the picture

taurus [48]3 years ago
3 0

Answer:

(-\frac{3}{4})(\frac{7}{8}) ↔ -\frac{21}{32}

(\frac{2}{3})(-4)(9) ↔ -24

(\frac{5}{16})(-2)(-4)(-\frac{4}{5}) ↔ -2

(2\frac{3}{5})(\frac{7}{9}) ↔ \frac{91}{45}

Step-by-step explanation:

The first expression is

(-\frac{3}{4})(\frac{7}{8})

On simplification we get

-\frac{3\times 7}{4\times 8}

-\frac{21}{32}

Therefore the product of (-\frac{3}{4})(\frac{7}{8}) is -\frac{21}{32}.

The second expression is

(\frac{2}{3})(-4)(9)

On simplification we get

(\frac{2}{3})(-36)

-\frac{72}{3}

-24

Therefore, the product of (\frac{2}{3})(-4)(9) is -24.

Similarly,

(\frac{5}{16})(-2)(-4)(-\frac{4}{5})\Rightarrow (\frac{5}{16})(8)(-\frac{4}{5})=(\frac{5}{2})(-\frac{4}{5})=-2

(2\frac{3}{5})(\frac{7}{9})=(\frac{13}{5})(\frac{7}{9})=\frac{91}{45}

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2

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Find the zero of each function and state the multiplicity of each zero. Please show all steps.
vodka [1.7K]

Answer:

1. y=(x+3)^3. Zero: x=-3 multiplicity 3.

2. y=(x-2)^2 (x-1). Zeros: x=2 multiplicity 2; x=1 multiplicity 1.

3. y=(2x+3)(x-1)^2. Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.


Step-by-step explanation:

1. y=(x+3)^3

y=0\\ (x+3)^3=0\\ \sqrt[3]{(x+3)^3}=\sqrt[3]{0}\\ x+3=0\\ x+3-3=0-3\\ x=-3

Zero: x=-3 multiplicity 3.


2. y=(x-2)^2 (x-1)

y=0\\ (x-2)^2(x-1)=0\\ \left \{ {{(x-2)^2=0} \atop {x-1=0}} \right\\ \left \{ {{\sqrt{(x-2)^2} =\sqrt{0} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x-2=0} \atop {x=1}} \right\\ \left \{ {{x-2+2=0+2} \atop {x=1}} \right\\ \left \{ {{x=2} \atop {x=1}} \right.

Zeros: x=2 multiplicity 2; x=1 multiplicity 1


3. y=(2x+3)(x-1)^2

y=0\\ (2x+3)(x-1)^2=0\\ \left \{ {{2x+3=0} \atop {(x-1)^2=0}} \right\\ \left \{ {{2x+3-3=0-3} \atop {\sqrt{(x-1)^2} =\sqrt{0} }} \right\\ \left \{ {{2x=-3} \atop {x-1=0}} \right\\ \left \{ {{\frac{2x}{2} =\frac{-3}{2} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x=-\frac{3}{2} } \atop {x=1}} \right.

Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.

4 0
4 years ago
Tell weather each equation has one, zero, or infinitely many solutions. -(2x + 2) - 1 = -x - (x + 3)
Anestetic [448]
-(2x+2)-1=-x-(x+3) \\
-2x-2-1=-x-x-3 \\
-2x-3=-2x-3 \\
0=0 \\
always \ true \\ \\
\boxed{\hbox{infinitely many solutions}}
7 0
3 years ago
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sineoko [7]

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5 0
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3 0
3 years ago
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