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sammy [17]
3 years ago
13

What is the greatest age that pat could be?

Mathematics
1 answer:
m_a_m_a [10]3 years ago
4 0
I dunno. Average human life expectancy is about 70-90, but other than that, I don’t know what you’re talking about, as you forgot to add pictures.
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Hi guys . pls help me with this​
daser333 [38]

Step-by-step explanation:

The upstream speed is S / t₁, and the downstream speed is S / t₂.

If we say f is the speed of the fish in calm water, and r is the speed of the river, then:

f − r = S / t₁

f + r = S / t₂

If we say T is the time it takes to cross the river, then the speed perpendicular to the river is ℓ/T, the speed parallel to the river is r, and the overall speed is f.

Using Pythagorean theorem:

f² = (ℓ/T)² + r²

f² − r² = (ℓ/T)²

(f − r) (f + r) = (ℓ/T)²

(S / t₁) (S / t₂) = (ℓ/T)²

S² / (t₁ t₂) = (ℓ/T)²

(t₁ t₂) / S² = (T/ℓ)²

√(t₁ t₂) / S = T/ℓ

T = ℓ√(t₁ t₂) / S

6 0
3 years ago
What is 0.2 in the 2 <br> keep going as a fraction
ollegr [7]
It is 2/10 for tour question
8 0
3 years ago
1. The graphs below represent functions defined by polynomials. For which function are the zeros of the
melamori03 [73]

Answer:

the second graph (c)

Step-by-step explanation:

8 0
3 years ago
Label the following as either function or relation ​
Katarina [22]

I can't see anything so I'll say function. Next time add a screenshot!

Best of luck to you

4 0
3 years ago
Find the multiplicative inverse of 6 + 2i
Marina CMI [18]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2774989

________________


Find the multiplicative inverse of

\mathsf{z=6+2i}

________


The inverse multiplicative of  \mathsf{z=a+bi}  is

\mathsf{\dfrac{1}{z}}\\\\\\&#10;=\mathsf{\dfrac{1}{a+bi}\qquad\quad(a\ne 0~~and~~b\ne 0)}\\\\\\&#10;=\mathsf{\dfrac{1}{a+bi}\cdot \dfrac{a-bi}{a-bi}}\\\\\\&#10;=\mathsf{\dfrac{1\cdot (a-bi)}{(a+bi)\cdot (a-bi)}}\\\\\\&#10;=\mathsf{\dfrac{a-bi}{a^2-\,\diagup\hspace{-10}abi+\,\diagup\hspace{-10}abi-(bi)^2}}

=\mathsf{\dfrac{a-bi}{a^2-b^2\cdot i^2}}\\\\\\&#10;=\mathsf{\dfrac{a-bi}{a^2-b^2\cdot (-1)}}\\\\\\&#10;=\mathsf{\dfrac{a-bi}{a^2+b^2}}\\\\\\\\&#10;\therefore~~\mathsf{\dfrac{1}{a+bi}=\dfrac{a}{a^2+b^2}-\dfrac{b}{a^2+b^2}\,i\qquad\quad\checkmark}

________


For this question,

\mathsf{z=6+2i}


So,

\mathsf{\dfrac{1}{z}}\\\\\\&#10;=\mathsf{\dfrac{1}{6+2i}}\\\\\\&#10;=\mathsf{\dfrac{6}{6^2+2^2}-\dfrac{2}{6^2+2^2}\,i}\\\\\\&#10;=\mathsf{\dfrac{6}{36+4}-\dfrac{2}{36+4}\,i}\\\\\\&#10;=\mathsf{\dfrac{6}{40}-\dfrac{2}{40}\,i}


\therefore~~\mathsf{\dfrac{1}{z}=\dfrac{3}{20}-\dfrac{1}{20}\,i}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)

6 0
3 years ago
Read 2 more answers
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