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2 years ago

9

Answer this question with a clear explanation to be marked as Brainiest ✨✨✨✨✨✨✨✨

1 answer:

Nina [5.8K]2 years ago

5 0

Answer:

C

Step-by-step explanation:

A, B and D are all only a few digits long and can be expressed as the ratio of two integers, where at least one is even:

A = -6/16 = -3/8, so it's rational

B = 7.0398 = 35199/5000, so it's rational

D = 6,907 = 13814/2, therefore rational

C, on the other hand is irrational.

If we simplify the surd √8, we get 2√2. The 2's on the top and bottom can now cancel out, giving us √2, which is irrational. [There's a proof but it's kinda long].

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Ashley increased the amount of water she drinks each day from 80oz to 87 oz.By what percentage did she increase her water intake

rewona [7]

87oz - 80oz = 7oz
7oz/80oz = 0.0875
0.0875 x 100% = 8.75%

Ashley increased her water intake by 8.75%.

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4 years ago

21. Create an area model to represent<br> 14 X 17. Find the product.

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3 years ago

The product of x and the sum of 6 and 8 times the square of x

tensa zangetsu [6.8K]

Answer:

x = 14; x = 0

Step-by-step explanation:

(x)(6 + 8) = x²

(x)(14) = x²

14x = x²

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3 years ago

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3 years ago

Suppose water is added to a tank at 10 gal / min , but leakes out at the rate of 0.2 gal / min for each gallon in the tank . Wha

Nana76 [90]

Answer:

50 gallon

Step-by-step explanation:

We are given that

$\frac{dV_{in}}{dt}=10 gal/min$

Let V(in gallon) be the present volume of tank.

$\frac{dV_{out}}{dt}=0.2 V gal/min$

We have to find the smallest capacity the tank can have if the process is to continue indefinitely.

According to question

$\frac{dV}{dt}$ =Incoming volume-outgoing volume

$\frac{dV}{dt}=10-0.2 V=-(\frac{1}{5}V-10)=-(\frac{V-50}{5})$

$\frac{5dV}{V-50}=-dt$

Integrating on both sides

$\int \frac{5dV}{V-50}=-\int dt$

$5ln(V-50)=-t+ln C$

By using the formula

$\int \frac{dx}{x}=ln x+C$

$ln(V-50)=\frac{1}{5}(-t+ln C)$

$ln(V-50)=-\frac{1}{5}t+\frac{1}{5} ln C$

$ln(V-50)=-\frac{1}{5} t+lnC^{\frac{1}{5}}$

$ln(V-50)=-\frac{1}{5}t+ln C'$

Where $ln C'=ln C^{\frac{1}{5}}$

$ln(V-50)-ln C'=-\frac{1}{5} t$

$ln\frac{V-50}{C'}=-\frac{1}{5}$

Using the formula

$ln m-ln n=ln\frac{m}{n}$

$\frac{V-50}{C'}=e^{-\frac{1}{5}t}$

$V-50=C'e^{-\frac{1}{5}t}$

Substitute $t=\infty$

$V-50=0$

$V=50$

Hence, the smallest capacity the tank can have if the process is to continue indefinitely=50 gallon

7 0

3 years ago