All categories

3 years ago

Helpppppppp

Mathematics

1 answer:

wel3 years ago

6 0

The hypotenuse can be solved by this formula. X^2 = 6^2 +6^2

X^2=64
Usually you cut it down to 8 but it wants the squared form so 64.

You might be interested in

What is 55 x 25 equaled to

nikitadnepr [17]

Answer:

1375

Step-by-step explanation:

8 0

3 years ago

Rewrite 3x+2y=6 in slope intercept form (it isn't y=-3/2x+3

arsen [322]

Answer:

according to my calculations (LOLLLLLLL) what you got is correct??? email ur teacher because i don't see anything wrong with your answer

Step-by-step explanation:

4 0

2 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

3 years ago

Zaid has a peculiar pair of four-sided dice. When he rolls the dice, the probability of any particular outcome is proportional t

Elden [556K]

Answer: a) $\bold{\dfrac{3}{16}}$ b) $\bold{\dfrac{1}{36}}$

<u>Step-by-step explanation:</u>

a) In order to get an even number, you have the 3 different scenarios:

1) Even, Even, Even, Even $\dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}$

2) Even, Even, Odd, Odd $\dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}$

3) Odd, Odd, Odd, Odd $\dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}$

<em>Order doesn't matter</em>

Add them up to get your answer: $\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}\quad =\large\boxed{\dfrac{3}{16}}$

b) If one die is a 2 and another is a 3 and the other two dice can be any number, then you have 1 possibility for a 2, 1 possibility for a 3, and 6 possibilities for each of the other two dice.

$\dfrac{1\times 1\times 6\times 6}{6^4}\quad =\dfrac{1}{6^2}\quad =\large\boxed{\dfrac{1}{36}}$

3 0

3 years ago

In a race, 27 out of the 50 finished in less than minutes. What percent of swimmers finished the race in less than 57 minutes?

alekssr [168]

Answer:

<u>27 x 2 </u> = <u>54 </u> = 54% 54% of racers finished in less than 57 minutes

50 x 2 = 100 =

7 0

2 years ago

Read 2 more answers