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Scorpion4ik [409]
3 years ago
10

How many solutions do this have 6(x+5)=6x+5

Mathematics
1 answer:
jok3333 [9.3K]3 years ago
3 0

Answer:

No solutions

Step-by-step explanation:

6(x+5) = 6x+5

6(x+5) → 6(x) + 6(5) → 6x + 30

6x + 30 = 6x + 5

–6x –6x

30 ≠ 5.

Try making x zero, and you will realize that 5 is never equal to 30 so there is no way to satisfy this equation.

In other words, there is a contradiction.

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If one pound is the same weight as 16 ounces, how many ounces equal 3.5 pounds? (Only input whole numbers)
siniylev [52]

Answer:

56

Step-by-step explanation:

16 x 3 = 48

Half of 16 is 8

48 + 8 = 56

4 0
3 years ago
Read 2 more answers
Solve the rational equation x divided by 6 equals x squared divided by quantity x minus 1 end quantity, and check for extraneous
Dmitriy789 [7]

Answer:

Solution: x=-\frac{1}{5},x=0

Step-by-step explanation:

Given:

The rational equation to solve is given as:

\frac{x}{6}=\frac{x^2}{x-1}

Doing cross product, we get:

x\times (x-1)=6\times x^2\\x^2-x=6x^2\\\textrm{Bringing all variables to the right side, we get:}\\6x^2-x^2+x=0\\5x^2+x=0\\x(5x+1)=0\\x=0\ or\ 5x+1=0\\x=0\ or\ x=-\frac{1}{5}

Now, for x=0, the rational equation is equal to 0. So, x=0 is a solution.

Also, x=-\frac{1}{5} is also a solution.

So, no extraneous solution.

4 0
4 years ago
Find the center of the circle that can be circumscribed about triangle EFG with E(2, 2), F(2, -2), and G(6,-2).
nadezda [96]
Hello friend that would be the last option (3, 2)
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6 0
3 years ago
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MZ3 = 64<br>m_4 = 106°<br>and<br>4<br>2<br>3<br>1<br>What is<br>mZ1,​
ddd [48]

Answer: m∠1=42°

Step-by-step explanation

For this problem, we can separate the triangle into 2 separate triangles to solve for m∠1.

We know that m∠4 is 106°. ∠2 and ∠4 are supplementary angles. You can see that they make a straight line of 180°. We can use this fo find ∠2.

∠2+∠4=180

∠2+106=180

∠2=74°

Now that we know ∠2 is 74°, we can use that to solve for m∠1.

We know that the measures of the angles in a triangle equal to 180°. For the triangle at the bottom, we see ∠3, ∠2, ∠1. Since we know ∠3 and ∠2, we can find ∠1.

∠3+∠2+∠1=180

64+74+∠1=180

138+∠1=180

∠1=42°

4 0
4 years ago
g Suppose that the populationy(t)of a certain kind of fish is given by the logistic modely′= 3y−2y2(a) Solve the Bernoulli equat
Ksju [112]

Answer:

y = \frac{3}{2 + 3Ce^{-3t} }

Step-by-step explanation:

As given , y' = 3y - 2y²

⇒y' - 3y = -2y²

Divide by y² in the above equation

⇒\frac{y'}{y^{2} } - \frac{3y}{y^{2} } = -\frac{2y^{2} }{y^{2} }

⇒\frac{y'}{y^{2} } - \frac{3}{y } = -2       ........(1)

Now , let \frac{1}{y} = u

⇒-\frac{1}{y^{2} } \frac{dy}{dt} = \frac{du}{dt}

⇒-\frac{1}{y^{2} }y' = \frac{du}{dt}

∴ equation (1) becomes

-\frac{du}{dt} - 3u = -2

⇒\frac{du}{dt} + 3u = 2

It is a linear differential equation

Now,

Integrating factor = I.F = e^{\int\limits {3} \, dt } = e^{3t}

∴ The solution becomes

u.(I.F) = \int\limits {2.(I.F)} \, dt + C

⇒u.(e^{3t}) = \int\limits {2.(e^{3t} )} \, dt + C

⇒u.(e^{3t}) = \frac{2e^{3t} }{3} + C

⇒u = \frac{2}{3} + Ce^{-3t}

As \frac{1}{y} = u

⇒\frac{1}{y}  = \frac{2}{3} + Ce^{-3t}

⇒ y = \frac{1}{\frac{2}{3} + Ce^{-3t}  } = \frac{3}{2 + 3Ce^{-3t} }

⇒ y = \frac{3}{2 + 3Ce^{-3t} }

6 0
3 years ago
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