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bazaltina [42]
3 years ago
5

How do I simplify the following 1-4(u-1)

Mathematics
1 answer:
lakkis [162]3 years ago
3 0
-4u+5
hope this helps you
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Write 7x-2-7x+6 in simplest form.
luda_lava [24]
7x - 7x = 0
-2 + 6 = 4

So, the answer is 4. 
5 0
3 years ago
A circle has a diameter with endpoints (-7, -1) and (-5, -9). What is the equation of the circle? r2 = (x + 6)2 + (y + 5)2 r2 =
Fantom [35]
Answer: (x + 6)² + (y + 5)² = 17

Explanation:

1)  The equation of a circle is: (x - a)² + (y - b)² = r², wnere:
a is the x-coordinate of the center of the circleb is the y-coordinate of the center of the circler is the radius of the circle
2) Since the points (-7,-1) and (-5,9) define a diameter, you can find the center of the circle as the mid point of the diameter:
x-coordinate of the mid point = (-7 - 5) / 2 = - 12 / 2 = - 6
y-coordinate of the mid point = (-9 - 1)/ 2 = - 10 / 2 = - 5

Therefore, the center of the circle is (-6,-5)

3) The radius of the center is half the diameter.
The length of the diameter is found using the formula for the distance between two points applied to the two endopoints given:
diameter² = (7 - 5)² + (- 1 + 9)² = 2² + 8² = 4 + 64 = 68

⇒diameter = √68 = 2√17

⇒ radius = diamter / 2 = 2√17 / 2 = √17

4) Now you can replace the coordinates of the center and the radius into the equation:
(x + 6)² + (y + 5)² = 17


8 0
3 years ago
an experiment consists of flipping a fair coin 4 times. What is the probability of obtaining at least one head?
Dvinal [7]
\mathbb P(X\ge1)=1-\mathbb P(X=0)

The probability of getting 0 heads in 4 tosses (or equivalently, 4 tails) is \dfrac1{2^4}=\dfrac1{16}.

So the desired probability is

1-\dfrac1{16}=\dfrac{15}{16}
7 0
3 years ago
A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so
Debora [2.8K]

Let A(t) = amount of salt (in pounds) in the tank at time t (in minutes). Then A(0) = 11.

Salt flows in at a rate

\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}

and flows out at a rate

\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}

which I'll solve with the integrating factor method.

\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95

-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}

\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}

Integrate both sides. By the fundamental theorem of calculus,

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)

\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}

\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}

\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}

After 1 hour = 60 minutes, the tank will contain

A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131

pounds of salt.

7 0
2 years ago
PLEASE HELPPP ME ITS DUE TODAY
mel-nik [20]

Answer:

Randall: 44, Amy: 35

Step-by-step explanation:

Four years ago, their age added up to 71. Since four years have passed and they've each grown four years older since then, their ages added up together is 79. Here is the equation for Amy: x + (x + 9) = 79. We can simplify to get 35. Now we add 9 to 35 to get Randall's age. So, Amy's age is 35 and Randall's age is 44, and 35 + 44 = 79.

3 0
3 years ago
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