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3 years ago

Solve. y=x - 7 5+2y=7

Mathematics

2 answers:

lesantik [10]3 years ago

8 0

Answer:

$\huge\boxed{x=8;\ y=1\to(8;\ 1)}$

Step-by-step explanation:

$\left\{\begin{array}{ccc}y=x-7&(1)\\5+2y=7&(2)\end{array}\right\\\\\text{Substitute (1) to (2) and solve for x:}\\\\5+2(x-7)=7\qquad\text{use the distributive property}\\\\5+(2)(x)+(2)(-7)=7\\\\5+2x-14=7\\\\2x+(5-14)=7\\\\2x-9=7\qquad\text{add 9 to both sides}\\\\2x-9+9=7+9\\\\2x=16\qquad\text{divide both sides by 2}\\\\\dfrac{2x}{2}=\dfrac{16}{2}\\\\\boxed{x=8}$

$\text{Substitute it to (1):}\\\\y=8-7\\\\\boxed{y=1}$

Illusion [34]3 years ago

4 0

Answer:

(8, 1).

Step-by-step explanation:

y = x - 7

5 + 2y = 7

5 + 2(y) = 7

5 + 2(x - 7) = 7

5 + 2x - 14 = 7

Subtract 5 from both sides

2x - 14 = 2

Add 14 to both sides

2x = 16

Divide both sides by 2

x = 8

Since x = 8...

y = x - 7

y = 8 - 7

y = 1.

To check our work...

5 + 2y = 7

5 + 2 * 1 = 7

5 + 2 = 7

7 = 7

Since it works out, your answer is (8, 1).

Hope this helps!

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3 years ago

Consider the function f(x)=x2e4x. f(x) has two inflection points at x = C and x = D with C≤D where C is and D is Finally for eac

fredd [130]

Answer:

the inflection points are

$x = -0.85, -0.14$

So,

$C = -0.85 \\D = -0.14$

It is concave down at the intervals

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And it would be concave up at

$(-0.85 , -0.14)$

Step-by-step explanation:

Remember that to find inflection points you need to find where

$f''(x) = 0$

Since

$f(x) = x^2 e^{4x}$

Then using the product and the chain rule you have that

$f'(x) = 2e^{4x} x(2x+1)$

And then, using again the chain rule and the product rule you have that

$f''(x) = 2e^{4x} (8x^2 +8x+1)$

Therefore you have to solve the equation

$8x^2 +8x+1 = 0$

Using the quadratic equation you get that there are two solutions, so the inflection points are

$x = -0.85, -0.14$

So,

$C = -0.85 \\D = -0.14$

Now remember that a function is concave up if the derivative is greater than zero and concave down if the derivative is less than zero. Therefor you have to solve these inequalties

$8x^2 +8x+1 < 0 \\8x^2 +8x+1 \geq 0$

And you would get that is concave down at the intervals

$(-\infty , -0.85] \cup [-0.14,\infty)$

And it would be concave up at

$(-0.85 , -0.14)$

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