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Neko [114]
3 years ago
6

The value of x in this system of equations is 1.

Mathematics
2 answers:
maria [59]3 years ago
5 0

<em>Answer:</em>

<em>In my opinion the answer is </em><em><u>y = 6</u></em>

<em>Step-by-step explanation:</em>

<em>We have two unknowns from the equation therefore, two equations are needed. These equations are: </em>

<em> </em>

<em>3x + y = 9  </em>

<em>y = –4x +10  </em>

<em> </em>

<em>To solve for y, we first substitute the second equation to the first one. </em>

<em>3x + –4x +10= 9  </em>

<em>x = 1 </em>

<em> </em>

<em>We substitute the value of x to either of the equations and solve for y. </em>

<em> </em>

<em>y = –4(1) +10  </em>

<em>y = 6</em>

castortr0y [4]3 years ago
5 0

Answer:

y=6

Step-by-step explanation:

3x + y = 9

y = -4x + 10

Substitute the value of y in the first equation:

3x + (-4x + 10) = 9

Apply the subtraction property of equality:

-x + 10 = 9

-x=-1

x=1

Substitute 1 for x in the given y equation

y = -4x + 10

y = -4(1)+ 10

y=6

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4 0
2 years ago
Four friends went to the movies each ticket cost $8 and each person bought popcorn and a soda for $5 how much did they spend in
Neporo4naja [7]
Answer: m = $52

Explanation:
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Using substitution, you plug each variable
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7 0
3 years ago
Find gradient <br><br>xe^y + 4 ln y = x² at (1, 1)​
cricket20 [7]

xe^y+4\ln y=x^2

Differentiate both sides with respect to <em>x</em>, assuming <em>y</em> = <em>y</em>(<em>x</em>).

\dfrac{\mathrm d(xe^y+4\ln y)}{\mathrm dx}=\dfrac{\mathrm d(x^2)}{\mathrm dx}

\dfrac{\mathrm d(xe^y)}{\mathrm dx}+\dfrac{\mathrm d(4\ln y)}{\mathrm dx}=2x

\dfrac{\mathrm d(x)}{\mathrm dx}e^y+x\dfrac{\mathrm d(e^y)}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x

e^y+xe^y\dfrac{\mathrm dy}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x

Solve for d<em>y</em>/d<em>x</em> :

e^y+\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x

\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x-e^y

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x-e^y}{xe^y+\frac4y}

If <em>y</em> ≠ 0, we can write

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2xy-ye^y}{xye^y+4}

At the point (1, 1), the derivative is

\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=1,y=1}=\boxed{\dfrac{2-e}{e+4}}

4 0
3 years ago
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