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Semenov [28]
3 years ago
15

Evaluate 15/r -1 when r = 5

Mathematics
2 answers:
Juli2301 [7.4K]3 years ago
4 0
\frac{15}{r}-1=\frac{15}{5}-1=3-1=2

2)
\frac{15}{r-1}=\frac{15}{5-1}=\frac{15}{4}=3\frac{3}{4}=3.75
Fynjy0 [20]3 years ago
4 0
15/r-1  
15/5-1  (substitute 5 for r)
15/4     (5-1 is 4)
3.75     (15/4 is 3.75)
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Step-by-step explanation:

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1 year ago
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.

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