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3 years ago

Which expression is equivalent to five to the tenth power multiplied by five to the fifth

Mathematics

2 answers:

ioda3 years ago

6 0

Answer:

30517578125

Step-by-step explanation:

do u mean multiplying it? if not then plz lmk so i could help

xxTIMURxx [149]3 years ago

4 0

Cacamole neidgeksyikeusjejdjdywisijss

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Every day The burger buns restaurant serves 1/5 of a bottle of cherry soda to the customers for how many days will 1 4/5 bottles

kogti [31]

9 because if you change 1 and 4/5 to an improper fraction and you’ll get 9/5. If you multiply 1/5 by 9 you’ll get 9/5.

5 0

2 years ago

Please help. Thank you.

Trava [24]

Answer:

By the Pythagoras theorem, the hypotenuse (c), is equivalent to:

a^2+ b^2= c^2 where a and b are the other two sides of the right angled triangle, hence c= square root (15^2+ 20^2)= 25

4 0

3 years ago

Two containers, one for 3 liters and one for 5 liters, how would you measure 4 liters

Mashutka [201]

Fill the 5-liter container, pour water from that into the 3 liter container until that is full, You will now have 2 liters left in the 5 liter container.

Empty the 3-liter container, and then transfer the 2 liters from the 5-liter container into it.

Now fill the 5-liter container again, then pour water carefully from the 5-liter container into the 3-liter container until it is full - exactly one more liter.

The 5-liter container now has 4 liters in it.

8 0

2 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago

Find the missing probability. P(B)=720,P(A|B)=14 ,P(A∩B)=?

Ivan

Answer:

7/80

Step-by-step explanation:

Given that: P(B) = 7 / 20, P(A|B)= 1 / 4

Bayes theorem is used to mathematically represent the conditional probability of an event A given B. According to Bayes theorem:

$P(A|B)=\frac{P(A \cap B)}{P(B)}$

Where P(B) is the probability of event B occurring, P(A ∩ B) is the probability of event A and event B occurring and P(A|B) is the probability of event A occurring given event B.

$P(A|B)=\frac{P(A \cap B)}{P(B)}\\\\P(A \cap B)=P(A|B)*P(B)\\\\Substituting:\\\\P(A \cap B)=1/4*7/20=7/80\\\\P(A \cap B)=7/80$

6 0

3 years ago