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3 years ago

Tell whether the ordered pair is a solution of the linear system

Mathematics

1 answer:

ANEK [815]3 years ago

6 0

3) Yes
4) No, the second equation ends up as 26=11, which is false
5) No, the second equation ends up as -4=0, which is false

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solong [7]

Answer:

she selling five dollars

Step-by-step explanation:

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3 years ago

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how do you solve the formula for an unknown variable in terms of the known variables? For context, the question deals with princ

pochemuha

The easiest literal equations to understand are formulas where one variable is expressed in terms of the others. To solve for a given variable, isolate that variable on one side of the equation so that it's solved in terms of the other variables. Make sure to use the same operations on both sides of the equation!

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3 years ago

Suppose a triangle has sides a,b, and c with side c the longest side, and that a^2+b^2>c^2. Let theta be the measure of the a

andreyandreev [35.5K]

Answer :

A the triangle is not a right triangle

D costheta <0

The law of cosines says:

a²=b²+c²−2bccos(θ)

Rewriting:

2bccos(θ)=b²+c²−a²<0

So cos(θ)<0cos(θ)<0 since 2bc>02bc>0. Since 0<θ<π0<θ<π in any triangle,

π/2<θ<π

So:

1. θ is not an acute angle.

2 The triangle is not a right triangle. In a right triangle, one of the angles is 90 degrees and the other two are then less than 90 degrees. This triangle has an angle greater than 90 degrees.

3. cos(θ)<0 is true.

4. cos(θ)>0 is false -3 says cos(θ) is negative; a number can't be both positive and negative.

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2 years ago

Find three consecutive integers such that the second is eight less than twice the third. only an algebraic solution will be exce

Sergio039 [100]

I think that it is 8, 10, 12

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3 years ago

1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat

brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

$x = 2$ is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) $f(x) = 6x + 6/x$

$f\prime(x) = 6 - 6/x^2 = 0$ and $x \neq 0$

So, the roots are $x = -1$ and $x = 1$

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum.

2) $f(x)=6-4/x+2/x^2$

$f\prime(x)=4/x^2-4/x^3=0$ and $x \neq 0$

So the root is $x = 1$

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum.

3) $f(x) = 8x^2/(x-4)$

$f\prime(x) = (8x^2-64x)/(x-4)^2=0$ and $x \neq 4$

So the roots are $x = 0$ and $x = 8$

The function is increasing on the interval (-∞,0] ∪ [8,∞)

The function is decreasing on the interval [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum.

4) $f(x)=6(x-2)^{2/3} +4=0$

$f\prime(x) = 4/(x-2)^{1/3}$ has no solution and $x = 2$ is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

$x = 2$ is absolute minimum.

5) $f(x)=8\sqrt x - 6x$ for $x>0$

$f\prime(x) = (4/\sqrt x)-6 = 0$

So the root is $x = 4/9$

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum.

5 0

3 years ago