Ask question

Ask question

All categories

2 years ago

9

Hey what's the answer !!

e=" \sqrt{98 - ( 17)} " alt=" \sqrt{98 - ( 17)} " align="absmiddle" class="latex-formula">

(ﾉ≧∇≦)ﾉ

2 answers:

xeze [42]2 years ago

7 0

Answer:

√98-17

=√ 81

=√ (9)^2

=9 is ans

Novosadov [1.4K]2 years ago

6 0

Answer:

$\sqrt{98 - ( 17)}$

${\sqrt{81}}$

${\sqrt{9²}}$ =±9

is your answer

You might be interested in

A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11

mixer [17]

Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're given that the tank initially holds $A(0)=100$ lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

$\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}$
$\implies A'(t)+\dfrac{11}{200+33t}A(t)=484$

Find the integrating factor:

$\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}$

Distribute $\mu(t)$ along both sides of the ODE:

$(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}$
$\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}$
$A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt$
$A(t)=22(200+33t)^{2/3}+C$

Since $A(0)=100$ , we get

$100=22(200)^{2/3}+C\implies C\approx-652.39$

so that the particular solution for $A(t)$ is

$A(t)=22(200+33t)^{2/3}-652.39$

The tank becomes full when the volume of solution in the tank at time $t$ is the same as the total volume of the tank:

$200+(44-11)t=500\implies 33t=300\implies t\approx9.09$

at which point the amount of salt in the solution would be

$A(9.09)\approx733.47\text{ lb}$

4 0

3 years ago

Find the product 3 2/5 and 5/8. Express answer in simplest form

Neko [114]

Answer:

2 1/8

Step-by-step explanation:

Hope this helps!!

7 0

3 years ago

Read 2 more answers

What is a perimeter in a regular pentagon 3in

damaskus [11]

3in x 5 sides = 15in
perimeter = 15in

6 0

2 years ago

The usher at a wedding asked each of the 80 guests whether they were a friend of the bride or of the groom. Find the probability

yawa3891 [41]

<u>Complete Question</u>

The usher at a wedding asked each of the 80 guests whether they were a friend of the bride or of the groom. The results are: 59 for Bride, 50 for Groom, 30 for both. Find the probability that a randomly selected person from this sample was a friend of the bride OR of the groom.

Answer:

0.9875

Step-by-step explanation:

Total Number of Guests which forms the Sample Space, n(S)=80

Let the Event (a friend of the bride) =A

Let the Event (a friend of the groom) =B

n(A) =59

n(B)=50

Friends of both bride and groom, $n(A \cap B)=30$

Therefore:

$n(A \cup B)=n(A)+n(B)-n(A \cap B)\\n(A \cup B)=59+50-30\\n(A \cup B)=79$

The number of Guests who was a friend of the bride OR of the groom = 79

Therefore:

The probability that a randomly selected person from this sample was a friend of the bride OR of the groom.

$P(A \cup B) =\dfrac{n(A \cup B) }{n(S)} \\\\=\dfrac{79 }{80}\\\\=0.9875$

4 0

3 years ago

I need help with this ASAP!!!!!!

ss7ja [257]

Answer:

Is that all the information? There seems to not be enough information in this. If this is all there is then I'll try my best, but if you have anything else then put it in the comments of this answer and I'll answer it from there

Step-by-step explanation:

May I get brainliest please? :)

3 0

3 years ago

Read 2 more answers