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hoa [83]
3 years ago
9

What would be the most logical first step for solving this quadratic equation?

Mathematics
1 answer:
Andru [333]3 years ago
8 0

Answer:

<u>C. Add 8 to both sides</u>

Step-by-step explanation:

This is done in order to equate the equation to zero so that it relates to the general equation.

{ax}^{2}  + bx + c = 0

c is (13 + 8)

You might be interested in
find two positive real numbers such that they sum to 108 and the product of the first times the square of the second is a maximu
topjm [15]

The two positive numbers are 36 and 72 which gives a sum equal to 108 and the product of 36 and the square of 72 is a maximum.

Any integer greater than zero is considered a positive number. A positive number can either be written as a number or with the "+" symbol in front of it.

Let us consider the two positive real numbers as x and y. Then, their sum is written as,

x+y=108

Then, y=108-x

And the product is written as,

P=xy²

Substitute value of y in the above equation, we get,

\begin{aligned}P&=x(108-x)^2\\&=x(11664-216x+x^2)\\&=11664x-216x^2+x^3\\&=x^3-216x^2+11664x\end{aligned}

Now, differentiate P with respect to x, and we get,

\begin{aligned}\frac{dP}{dx}&=3x^2-432x+11664\\&=x^2-144x+3888\end{aligned}

Solving the above equation to zero, we get,

\begin{aligned}x^2-144x+3888&=0\\x^2-36x-108x+3888&=0\\x(x-36)-108(x-36)&=0\\(x-108)(x-36)&=0\\x&=\text{108 or 36}\end{aligned}

Substitute values of x in y=108-x, to get values of y.

If we substitute x=108, we get the y value as zero which doesn't give the required solution.

But, if we substitute x=36, we get,

\begin{aligned}y&=108-36\\y&=72\end{aligned}

Thus, the two positive numbers are 36 and 72.

To know more about positive numbers:

brainly.com/question/1635103

#SPJ4

4 0
1 year ago
(3√7 +5)^2 <br> simplify the problem to its simplest form.
klasskru [66]

Answer:

88 + 30\sqrt{7} \\

Step-by-step explanation:

Expand to make easier:

(3\sqrt{7} \\ + 5) x (3\sqrt{7} \\ + 5)

FOIL:

9(7) + 15    \sqrt{7} + 15\sqrt{7} + 25

Simplify:

63 + 30\sqrt{7} + 25

88 + 30\sqrt{7}

5 0
3 years ago
Find three consecutive integers such that the sum of the largest and 5 times the smallest is -244. URGENT
Bad White [126]

Answer:

-39+5(-41)=244

Step-by-step explanation:

Consider 3 consecutive integers: (smallest) x, x+1 and x+2 (largest)

Five times the smallest means 5x

Sum of the largest and five times the smallest means x+2+5x

This sum is -244 means x+2+5x=-244

Solve this equation

x+2+5x=-244

6x+2=-244

6x=-244-2

6x=-246

x=-41

So, smallest number is -41, second is -40 and largest is -39

= -39+5(-41)=244

Hope this helps, have a nice day/night! :D

5 0
3 years ago
What is the inverse of f(x)=x^3-2
lana66690 [7]
Would it be X=Y^3-2 bc I think you flip the x and y
7 0
4 years ago
Factorise Fully x^2 + x
balu736 [363]

Answer:

X(x+1)

Step-by-step explanation:

If it is wrong I'm sorry.

5 0
3 years ago
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