If x < -4, then -4 is a solution.<br> True<br> False

2x-4y=8 and 7x-3y=61 solve by substitution

Ratling [72]

Answer:

(10, 3)

Step-by-step explanation:

Solve by Substitution

2x − 4y = 8 and 7x − 3y = 61

Solve for x in the first equation.

x = 4 + 2y 7x − 3y = 61

Replace all occurrences of x with 4 + 2y in each e quation.

Replace all occurrences of x in 7x − 3y = 61 with 4 + 2y. 7 (4 + 2y) − 3y = 61

x = 4 + 2y

Simplify 7 (4 + 2y) − 3y.

28 + 11y = 61

x = 4 + 2y

Solve for y in the first equation.

Move all terms not containing y to the right side of the equation.

11y = 33

x = 4 + 2y

Divide each term by 11 and simplify.

y = 3

x = 4 + 2y

Replace all occurrences of y with 3 in each equation.

Replace all occurrences of y in x = 4 + 2y with 3. x = 4 + 2 (3)

y = 3

Simplify 4 + 2 (3).

x = 10

y = 3

The solution to the system is the complete set of ordered pairs that are valid solutions.

(10, 3)

The result can be shown in multiple forms.

Point Form:

(10, 3)

Equation Form:

x = 10, y = 3

4 0

3 years ago

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True or false. the function is the product of 400 and a factor that does not depend on the initial number of visitors

Aleks [24]

Answer:

true

Step-by-step explanation:

8 0

3 years ago

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Factor completely x^8- 16.

Nataly_w [17]

Answer:

Step-by-step explanation:

So in this example we'll be using the difference of squares which essentially states that: $(a-b)(a+b)=a^2-b^2$ or another way to think of it would be: $a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})$ . So in this example you'll notice both terms are perfect squares. in fact x^n is a perfect square as long as n is even. This is because if it's even it can be split into two groups evenly for example, in this case we have x^8. so the square root is x^4 because you can split this up into (x * x * x * x) * (x * x * x * x) = x^8. Two groups with equal value multiplying to get x^8, that's what the square root is. So using these we can rewrite the equation as:

$x^8-16 = (x^4-4)(x^4+4)$

Now in this case you'll notice the degree is still even (it's 4) and the 4 is also a perfect square, and it's a difference of squares in one of the factors, so it can further be rewritten:

$x^4-4 = (x^2-2)(x^2+2)$

So completely factored form is: $(x^2-2)(x^2+4)(x^4+4)$

I'm assuming that's considered completely factored but you can technically factor it further. While the identity difference of squares technically only applies to difference of squares, it can also be used on the sum of squares, but you need to use imaginary numbers. Because $x^2+4 = x^2-(-4)$ . and in this case a=x^2 and b=-4. So rewriting it as the difference of squares becomes: $x^4+4 = x^4 - (-4) = (x^2-\sqrt{-4})(x^2+\sqrt{-4}) = (x^2-2i)(x^2+2i)$ just something that might be useful in some cases.