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2 years ago

Someone Help me please

Mathematics

2 answers:

Slav-nsk [51]2 years ago

8 0

Answer:

35 degrees

Step-by-step explanation:

pav-90 [236]2 years ago

8 0

The answer is 35 degrees

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Use f(x) = 2 x and f-'(x) = 2x to solve the problems.

Marrrta [24]

Answer:

a) 1

b) 2

c) 2

Step-by-step explanation:

a) $f(2)$ means find the output for $\frac{1}{2}x$ when $x=2$ .

$\frac{1}{2}(2)$

$\frac{2}{2}$

$1$

b) $f^{-1}(1)$ means find the output for $2x$ when $x=1$

$2(1)$

$2$

c) If $f$ and $f^{-1}$ are truly inverses then $f(f^{-1}(u))=u$ and $f^{-1}(f(u))=u$ as long as $u$ satisfies the domains.

So we should be able to conclude that $f^{-1}(f(2))=2$ with no work.

However, I will also show work.

$f^{-1}(f(2))$

$f^{-1}(1)$ since $f(2)=1$ from part a.

$2$ since $f^{-1}(1)=2$ from part b.

6 0

3 years ago

Apply each of the four counting/sampling methods (with replacement and with ordering, without replacement and with ordering, wit

dusya [7]

Answer:

Step-by-step explanation:

I will illustrate this solution with a unique birthday party situation between Jane (the celebrant) and her 10 friends.

In the said party , we will assume that Jane only has 5 chocolates to share among her friends

i. Assuming that the 5 chocolates are of the same type, if she doesn't want to give any friend more than one piece of chocolate, the situation here is said to be WITHOUT REPLACEMENT & UN-ORDERED

n = 10 and k = 5

Total number of possible combinations becomes,

$(\left {n} \atop {k}} \right. )=(\left {10} \atop {5}} \right. )=252$

ii. Assuming that the 5 pieces of chocolate are of the same type and she is willing to give a friend more than one piece of chocolate, the situation is said to be WITH REPLACEMENT & UN-ORDERED

n = 10 and k = 5

Total number of possible combinations becomes,

$(\left {n+k-1} \atop {k}} \right. )=(\left {14} \atop {5}} \right. )=2002$

iii. Assuming that the 5 pieces of chocolate are of different types and she isn't willing to give any friend more than one piece of chocolate, the situation is said to be WITHOUT REPLACEMENT & ORDERED

n = 10 & k = 5

Total number of possible combinations becomes,

$P\left {n} \atop {k}} \right=P\left {10} \atop {5}} \right. =30240$

iv. Assuming the the 5 pieces of chocolate are of different types, if she is willing to give a friend more than one piece of chocolate, the situation is said to be WITH REPLACEMENT & ORDERED

n = 10 AND k = 5

Total number of combinations becomes

$n^k=10^5=100,000$