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3 years ago

If 2 times an integer x is increased by 7, the result is always greater than 16 but less than 29.

Mathematics

1 answer:

Degger [83]3 years ago

3 0

Answer:

X=6

Step-by-step explanation:

Turning the question into a formulae

2X + 5 > 16; and 2X + 5 < 29

Let’s start by calculating the lower limit

2X + 5 = 16

X = (16–5)/2

X = 11/2 = 5.5

X > 5.5

The lowest value of X is infinitesimally close to 5.5, but greater than 5.5.

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nasty-shy [4]

Answer:

The value of x is 68, all the angles in a parrelelogram equal 360, 68 + 112 + 112 + x = 360, all the angles are 292, and subtracted from 360 you get 68

Step-by-step explanation:

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3 years ago

Question 11 point(s) Jared made Ask Your Teacher cups of snack mix for a party. His guests ate Ask Your Teacher of the mix. How

liq [111]

Answer:

$8\frac{1}{2}$ cups

Step-by-step explanation:

To determine the quantity of snack mix Jared guest ate, we multiply the quantity of snack mix the guest ate by the quantity of snack mix Jared made. i.e we determine $\frac{2}{3}of12\frac{3}{4}$ by multiplying both fractions together

quantity of snack mix Jared guest ate = $\frac{2}{3}of12\frac{3}{4}=\frac{2}{3}*12\frac{3}{4}=\frac{2}{3}*\frac{51}{4}=\frac{17}{2} =8\frac{1}{2}$

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3 years ago

What is seventy-nine divided by twenty-five.

Elodia [21]

3.16 is the correct answer

4 0

3 years ago

Read 2 more answers

The difference of two positive numbers is 7 and the sum of their squares is 109. Find the numbers.

Ray Of Light [21]

Answer:

$Numbers\ are\ 58\ and\ 51.$

Step-by-step explanation:

The the numbers are $x\ and\ y$ .

Difference:

$Difference\ of\ numbers=7\\\\x-y=7....................eq(1)$

Sum:

$Sum\ of\ numbers=109\\\\x+y=109...........................eq(2)$

Solve $eq(1)$ and $eq(2)$ .

$eq(1)+eq(2)\\\\x-y+x+y=7+109\\\\2x=116\\\\x=58\\\\Now\ from\ equation\ 1\\\\x-y=7\\\\y=x-7\\\\y=58-7\\\\y=51$

7 0

3 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

aliya0001 [1]

Answer:

$A=1500-1450e^{-\dfrac{t}{250}}$

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

$R_{in}$ =(concentration of salt in inflow)(input rate of brine)

$=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}$

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration, $C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}$

$R_{out}$ =(concentration of salt in outflow)(output rate of brine)

$=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}$

Now, the rate of change of the amount of salt in the tank

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{250}$

We solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}$

Recall that when t=0, A(t)=50 (our initial condition)

$50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}$

4 0

3 years ago