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3 years ago

HELP ME 10 POINTS ASAP

Mathematics

2 answers:

choli [55]3 years ago

8 0

Answer:

12i√5 Hope this is right. I took the quiz so it shouldn't be wrong.

ioda3 years ago

3 0

Answer:

−

sin

(

)

⋅

cot

(

)

Step-by-step explanation:

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D all the time and I like the time you

andriy [413]

Answer:

Step-by-step explanation:

1a+1b(a+b-c)+1b+1c(b=c-a)+ 1a+1c(c+a-b)

1a+1b-c+1b +1c-a+1a+1c-b

1a+2b-c+a+1c-b

2a+1b+c this is the answer I think lol

6 0

2 years ago

Can someone please help me answer this Math problem?

kap26 [50]

Answer:

2√2.

Step-by-step explanation:

We use the Pythagoras Theorem:

x^2 = 2^2 + 2^2

x^2 = 8

x = √8

= 2√2.

4 0

3 years ago

Mike had a 16in by 20in photo he cropped 3in off of each side of the photo what are the perimeter and area of the photo

mihalych1998 [28]

Answer: perimeter: 30 in. Area: 221 in

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The sum of consecutive integers 1,2,3,...,n is given by the formula 1/2n(n+1). How many consecutive integers, starting with 1, m

LuckyWell [14K]

1/2 * n * (n+1) = 1081

distribute

1/2 n^2 +1/2n -1081=0

factor out 1/2

1/2 (n^2 +n -2162) =0

multiply each side by 2

(n^2 +n -2162) =0

factor

(n-46) (n+47) =0

n=46, n=-47

n cannot be negative

n=46

Answer 46

7 0

3 years ago

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

3 0

3 years ago