$\underbrace{6400}_{\underbrace{64}_{\underbrace{8}_{\underbrace{4}_{\boxed{2*2}} \boxed{2}}*\underbrace{8}_{\underbrace{4}_{\boxed{2*2}}\boxed{2}}}*\underbrace{100}_{\underbrace{10}_{\boxed{5*2}}*\underbrace{10}_{\boxed{5*2}}}}$

<u>Looking at our factors, the prime factorization of 6400 is $2^8*5^2$ </u>

5 0

3 years ago

Find x when y=5 if y varies inversely as x and x=6 when y =-18

Vera_Pavlovna [14]

It has been given that y varies inversely with x, therefore the equation for this relationship is as follows;
y = $\frac{k}{x}$ \
when y = -18 and x = 6
-18 = k/ 6 ---1)
We need to find x when y = 5
5 = k/x
when we divide 1)/2), k gets cancelled

$\frac{-18}{5}= \frac{x}{6}$
x = -18*6/5
x = -21.6
when y = 5, then x = -21.6

7 0

3 years ago

5 1⁄5 decimeters = centimeters

blsea [12.9K]

Answer:

$\large\boxed{5\dfrac{1}{5}\ decimeters=52\ centimeters}$

Step-by-step explanation:

$1\ dm=10\ cm\qquad\text{(1 decimeters = 10 centimeters)}\\\\\text{Therefore}\\\\5\dfrac{1}{5}\ dm=\dfrac{5\cdot5+1}{5}\ dm=\dfrac{26}{5}\ dm=\dfrac{26}{5}\cdot10\ cm=\dfrac{260}{5}\ cm=52\ cm$

5 0

3 years ago

The water level of a certain body of water is changing at a rate of W(t)=3/4cos(4-t/2) inches per hour, where t represents hours

Leni [432]

Answer:

The answer is below

Step-by-step explanation:

a) The integral $\int\limits^{10}_{1 } {W(t)} \, dt$ means the total change in level of water in inches from 1 am to 10 am. That is the water level change measured in inches between 1 am and 10 am

b) 7 pm is represented as t = 19 hours, while midnight is represented as t = 24 hours, hence the water level change between 7 pm and midnight is:

$\int\limits^{24}_{19 } {W(t)} \, dt\\\\$

c) Using casio calculator to evaluate the integral gives:

$\int\limits^{24}_{19 } {W(t)} \, dt\\\\$ = 2.54 inches

D) The total daily change in water level = $\int\limits^{24}_{0 } {W(t)} \, dt\\\\$ = 0.35 inches

8 0

3 years ago