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3 years ago

How do you solve x^2 plus 16x plus 63=0?

1 answer:

Alex3 years ago

8 0

$x^2+ 16x + 63=0 \\ \\a=1, \ b=16 , \ c=63 \\ \\ \Delta =b^2-4ac =16^2 -4\cdot1\cdot 63 = 256 -252 = 4 \\ \\x_{1}=\frac{-b-\Delta }{2a}=\frac{-16-\sqrt{4}}{2 }=\frac{-16-2}{2}=-\frac{-18}{2}=-9 \\ \\x_{2}=\frac{-b+\Delta }{2a}=\frac{-16+\sqrt{4}}{2 }=\frac{-16+2}{2}=-\frac{-14}{2}=-7 \\ \\ \\Answer : \ x= -7 \ \ or \ \ x= -9$

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What is the common ratio of the sequence?

Semenov [28]

Answer:

A -3

Step-by-step explanation:

-2 x -3 = 6

6 x -3 = -18

-18x -3 = 54

5 0

3 years ago

I need help ASAP. I need help can someone explain and can anyone tell me the answer please . I’ll give u anything u want

AleksAgata [21]

<h3>Answers:</h3>

Domain: $(-\infty, \infty)$ which is the set of all real numbers
Range: $[3, \infty)$ which is the set of y values such that $y \ge 3$
x intercepts: none
y intercept: only one y intercept and it is at (0,7) this is ordered pair notation and not interval notation
Interval positive: $(-\infty, \infty)$
Interval negative: none
Interval increasing: $(-2, \infty)$
Interval decreasing: $(-\infty, -2)$
Average rate of change over [-2,0] is 2

=====================================================

Explanation:

The domain is the set of all real numbers. This is because the graph goes on forever to the left and right. We can use any x value we want as an input. There are no restrictions to worry about such as to prevent dividing by zero errors.

To say "all real numbers" in interval notation, we write $(-\infty, \infty)$ which is another way of saying $-\infty < x < \infty$

-------------------

The range in interval notation is $[3, \infty)$ since y = 3 is the smallest y output possible, and we could have larger y values as well. So basically $y \ge 3$ can be used to describe the range without saying much else.

Note the use of a square bracket to include 3 as part of the interval.

-------------------

There are no x intercepts because this graph does not cross the x axis. The lowest point is at (-2,3) so there's no way we could reach y = 0. Put another way, y = 0 is not part of the range so we cannot have any x intercepts.

There is one y intercept and it is at (0,7) where the graph crosses the y axis. For any function, the max number of y intercepts is 1.

-------------------

When it says "interval positive", its asking "which part(s) of the graph are above the x axis?". That would be the entire graph meaning we have the interval $(-\infty, \infty)$ . Every point on this V shaped curve is of the form (x,y) where y is positive.

So this means that we do not have any points with a negative y value, and therefore the answer to "interval negative" is none.

-------------------

Now onto the "interval increasing". This is similar to the previous section, but now we're looking when the graph is going uphill as we read from left to right. This happens on the interval $(-2, \infty)$ or put another way when x > -2.

The graph goes downhill whenever x < -2. So that's why the answer for the "interval decreasing" is $(-\infty, -2)$

-------------------

Note the points (-2,3) and (0,7) are on the V shaped graph. These points have x coordinates of x = -2 and x = 0, which are the endpoints of the interval we're focusing on.

Compute the slope of the line through those two points

m = (y2-y1)/(x2-x1)

m = (7-3)/(0-(-2))

m = (7-3)/(0+2)

m = 4/2

m = 2

The positive slope means the line goes uphill as we read from left to right

The average rate of change on the interval [-2, 0] is the value 2

In other words, we go up 2 units each time we move to the right 1 unit.

8 0

3 years ago

A shopkeeper allows 15% discount on the marked price, still he manages to have 7% profit. How much high did he mark his goods ab

lianna [129]

Answer: $25.9\%$

Step-by-step explanation:

Given

Shopkeeper allows 15% discount on the marked price and still manages a profit of 7%

Suppose the marked price is $x$

So, the selling price is $(1-0.15)x=0.85x$

Suppose the cost price is $y$

$\Rightarrow \dfrac{0.85x-y}{y}=7\%\\\\\Rightarrow \dfrac{0.85x}{y}-1=0.07\\\\\Rightarrow \dfrac{0.85x}{y}=1.07\\\\\Rightarrow y=\dfrac{0.85x}{1.07}\\\\\Rightarrow y=0.794x$

So, the percentage the shopkeeper marked his goods above cost price

$\Rightarrow \dfrac{x-y}{y}\times 100\\\\\Rightarrow \dfrac{x-0.794x}{0.794}\times 100\\\\\Rightarrow \dfrac{0.2056}{0.794x}\times 100\\\\\Rightarrow 25.89\%\approx 25.9\%$