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3 years ago

Three times a number, x, increased by four is equal to five times the number, x. Which equation can be used to solve for x?

Mathematics

2 answers:

sesenic [268]3 years ago

5 0

Answer:

X = 2

Step-by-step explanation:

From the question we can gather some pretty important information. For starters, "three times a number x" would give us 3x. Secondly, "increased by 4" would give us +4. Lastly, "equal to five times x" would be = 5x. If we put this all together we have the following Algebraic Equation.

$3x+4 = 5x$

Now we can start solving the equation.

$3x+4 = 5x$

$4 = 2x$

$2 = x$

So we can see that the value of x in this equation is 2.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

Dimas [21]3 years ago

4 0

3x+4=5x
4=5x-3x
4=2x
X=2

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In-s [12.5K]

Answer:

$\frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}$

Step-by-step explanation:

Let L be the length of the ladder,

Given,

x = the distance from the base of the ladder to the wall, and t be time.

y = distance from the base of the ladder to the wall,

So, by the Pythagoras theorem,

$L^2 = y^2 + x^2$

$\implies L = \sqrt{y^2 + x^2}$ ,

Differentiating with respect to time (t),

$\frac{dL}{dt}=\frac{d}{dt}(\sqrt{x^2 + y^2})$

$=\frac{1}{2\sqrt{x^2 + y^2}}\frac{d}{dt}(x^2 + y^2)$

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$=\frac{1}{\sqrt{x^2 +y^2}}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Here,

$\frac{dy}{dt}=-6\text{ ft per sec}$

Also, $\frac{dL}{dt} = 0$ ( Ladder length = constant ),

$\implies \frac{1}{\sqrt{x^2 +y^2}}(x(-6)+y\frac{dy}{dt})=0$

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Which is the required notation.

8 0

3 years ago

Pre-Calc :

nikitadnepr [17]

The answer is
x=2t and t=x/2, but we have y = 3t -1
so y = 3(x/2) -1
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snow_tiger [21]

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Mark as brainliest if you find it helpful.

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jonny [76]

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3 years ago

4(x-6)<-2x+6 what is the solution to the inequality _

Cloud [144]

Simplify both sides if needed. The left-hand side needs simplification.

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All is left to do is add and subtract to get the x variable all alone.

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