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3 years ago

6x+6y=-6, 5x+y=-13Show your process pleaseSolve the linear equation

Mathematics

1 answer:

lisabon 2012 [21]3 years ago

5 0

Answer: Rewrite equations:

5x+y=−13;6x+6y=−6

Step: Solve5x+y=−13for y:

5x+y=−13

5x+y+−5x=−13+−5x(Add -5x to both sides)

y=−5x−13

Step: Substitute−5x−13foryin6x+6y=−6:

6x+6y=−6

6x+6(−5x−13)=−6

−24x−78=−6(Simplify both sides of the equation)

−24x−78+78=−6+78(Add 78 to both sides)

−24x=72

−24x

−24

(Divide both sides by -24)

x=−3

Step: Substitute−3forxiny=−5x−13:

y=−5x−13

y=(−5)(−3)−13

y=2(Simplify both sides of the equation)

Answer:

x=−3 and y=2

Step-by-step explanation:

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4 years ago

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Vedmedyk [2.9K]

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Step-by-step explanation:

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4 years ago

Which of the following statements are true about the graph of f(x) = -2tan(3x - 6) + 1? Select all that apply.

Digiron [165]

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3 years ago

<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B

riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

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2 years ago

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mina [271]

Answer

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Step-by-step explanation:

Check attachment

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3 years ago

6x+6y=-6, 5x+y=-13Show your process pleaseSolve the linear equation​

6x+6y=-6, 5x+y=-13Show your process pleaseSolve the linear equation