H(x)=(x^2)+1 and K(x)=-(x^2)+4. If K(H)=0, what are the roots/solutions?a. ±i√3
b. ±i
c. ±2
d. ±1
e. ±1, ±i√3

Answers

Answer 1

Answer: $H(x)=(x^2)+1\ \ \ \ and\ \ \ \ K(x)=-(x^2)+4. \n\nK(H)=-(H^2)+4=-(x^2+1)^2+4=4-(x^2+1)^2=\n\n.\ \ \ =2^2-(x^2+1)^2=(2-x^2-1)(2+x^2+1)=(1-x^2)(3+x^2)=\n\n.\ \ \ =(1-x)(1+x)(x^2-3\cdot i^2)=(1-x)(1+x)(x- √(3) \cdot i)(x+ √(3) \cdot i)\n\n K(H)=0\ \ \ \ \Leftrightarrow\ \ \ (1-x)(1+x)(x- √(3) \cdot i)(x+ √(3) \cdot i)=0\n\nx=1\ \ \ \ or\ \ \ \ x=-1\ \ \ \ or\ \ \ \ x=√(3) \cdot i\ \ \ \ or\ \ \ \ x=-√(3) \cdot i\n\nAns.\ e.$

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Can you solve (4x-3)/(x+2)+(15)/(x-3)

Answers

Please see the pic, I'd solved in it.

$(4x-3)/(x+2)$ + $(15)/(x-3)$ = 0
1st step: We need to have a common denominator.
So we do this:
$((4x-3)*(x-3))/((x+2)*(x-3))$ + $((15)*(x+2))/((x-3)*(x+2))$ = 0
2nd step: Distribute them
$(4x^(2)-3x-12x+9)/((x-3)(x+2))$ + $(15x+30)/((x+2)(x-3))$ = 0
3rd step: Do the rest of the equation!
We can put it on one fraction, its fine.
$(4x^2-15x+9+15x+30)/((x+2)(x-3))$ = 0
$(4x^2+39)/((x+2)(x-3))$ = 0 (-15 and 15 cancels out)
4th step: take the denominator to the other side
4x²+39 = 0 (anything × 0 = 0)
5th step: Take 39 to the other side
4x² = -39
6th step: Divide by 4 on both sides
$(4x^2)/(4)$ = $-(39)/(4)$
(4 and 4 cancels out)
x² = $-(39)/(4)$
7th step: Take the square root to keep x alone
x = $-\sqrt(39)/(4)$
It can also be written as:
x = $-( √(39) )/( √(4) )$
Final answer:
x = $-( √(39) )/(2)$

George rides his bike 9km south and then 12 km east. How far is he from his staring point?

Answers

15 km hopes helps u...

Which is bigger 0.125 or 0.15

Answers

0.125 is smaller, and 0.15 is bigger.

Here's how u tell:
The largest place in a decimal is the tens place, first u check the tens place of the decimals. OK, they are both 1. Then you move to hundredths place, see here is where's the catch. One is 2, and one is 5. So you go for the one with the five bc thats bigger.

so, 0.15 is bigger

0.125 is smaller than 0.150.150=0.150.150-0.125=0.025 there is a difference0.125-0.150= - number

(01.02) The number of roses (r) in a garden increases when the number of weeds (w) decreases. Write the correct equation for this scenario, and solve for the number of roses when there are 5 weeds.

Answers

Answer:

-1w=1r

If there are 5 fewer weeds, there will be 5 more roses

Step-by-step explanation:

The question states that the roses increase when the weeds decrease.

If an object is dropped from a height of 144 feet, the function h(t) = -16t2 + 144 gives the height of the object after t seconds. When will the object hit the ground? A) 9 seconds

B) 6 seconds

C) 1.5 seconds

D) 3 seconds

Answers

So this question is asking for the zeros, or the x-intercepts, of this equation. For this, we have to set h(t) to zero to solve for it.

For this, we will be using the quadratic formula, which is $x=(-b+√(b^2-4ac))/(2a),(-b-√(b^2-4ac))/(2a)$ (a = t^2 coefficient, b = t coefficient, and c = constant). In this case, our equation is $t=(-0+√(0^2-4*(-16)*144))/(2*(-16)),(-0-√(0^2-4*(-16)*144))/(2*(-16))$ . From here we can solve for t.

Firstly, solve the exponents and multiplications: $t=(-0+√(0+9216))/(-32),(-0-√(0+9216))/(-32)$
Next, solve the additions and subtractions: $t=(√(9216))/(-32),(-√(9216))/(-32)$
Next, square root 9216: $t=(96)/(-32),(-96)/(-32)$
Lastly, divide and your answers are going to be $t=-3,3$