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mafiozo [28]
3 years ago
8

Let f(x)=4-x^2, g(x) =2-x find (f+g)(x) and its domain

Mathematics
1 answer:
boyakko [2]3 years ago
8 0
Add like terms together..
-x^2 -x+6
............
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Solve this problem to find Y. 800 x Y = 4800
kupik [55]

Answer:

6

Step-by-step explanation:

800*Y=4800

Y=4800/800

Y=6

5 0
3 years ago
Read 2 more answers
Which rays are part of line BE?<br><br> AC and AE<br> AB and AE<br> AC and AB<br> AB and
rosijanka [135]

We are given a diagram with different rays and lines.

Note: A line can be extend more and more upto infinitely length from both of the ends but a ray can be extend from from end only.

We need to find the rays for the line BE.

We can see BE line has a point A in between line.

So, we can name the rays formed with end point A as AB and AE.

<h3>Therefore,  AB and AE rays are part of line BE.</h3><h3>So, the correct option is 2nd option :</h3><h3>AB and AE.</h3>
4 0
3 years ago
Read 2 more answers
Azul spun a tri-color spinner twice. What is the probability that the spinner landed on the colors blue and green in any order?
guapka [62]

Answer:

The Probability that the spinner landed on the colors blue and green in any order = 2/9

Step-by-step explanation:

Given - Azul spun a tri-color spinner twice.

To find - What is the probability that the spinner landed on the colors blue and green in any order?

Solution -

Given that,

A tri-color spinner spun twice

So,

The Sample Space, S = {BB, BG, BY, GB, GG, GY, YB, YG, YY}

⇒n(S) = 9

Now,

Let A be an event that the spinner landed on the colors blue and green

So,

A = {BG, GB}

⇒n(A) = 2

Now,

Probability that the spinner landed on the colors blue and green in any order = n(A) ÷ n(S)

         = 2 ÷ 9

∴ we get

The Probability that the spinner landed on the colors blue and green in any order = 2/9

8 0
3 years ago
Given the functions below, find (g * h)(1). g(x) = x^2 + 4 + 2x. h(x) = -3x + 2
pshichka [43]

Answer:

3

Step-by-step explanation:

g(x) = x² + 2x+4

h(x) = -3x+2

(g*h)(1) is the same as

g(h(1)) , next solve for h(1) first by substituting in h(x), x with 1

g( h( x= 1)) = g( -3*1 +2)  = g( -1) so substitute in g(x) , x with -1

g(x= -1) = (-1)² +2(-1) +4 =1-2+4 =3

7 0
3 years ago
The line y = hx - 2 where h &gt; 0 meets the curve 3y² = 9x² - 6x + 1.
I am Lyosha [343]

Answer:

h=6

Step-by-step explanation:

since y =hx -2 is an equation for a line which intersects with the curve 3y^2 = 9x^2 - 6x +1. The point of intersection, let's say (x_1,y_1), should satisfy the two equations. As a result, the value of y in the second equation can be replaced with the value of y in the first equation as the following,

3y^2 = 3(hx-2)^2 = 9x^2 - 6x +1\\3(h^2x^2 -4hx + 4) = 9x^2 - 6x + 1\\3h^2 x^2 -12hx +12 = 9x^2 - 6x +1

therefore, the latter equation can be rewritten in a quadratic equation form as the following,

(9 - 3h^2) x^2 + (12h-6)x + (1-12) = (9 - 3h^2) x^2 + (12h-6)x -11 = 0

if the line is tangent to the curve, it means that the line touches the curve at one point, therefore the discernment of the second order equation will be equal to zero for the famous quadratic equation solution.

b^2 -4ac =0

where a=9-3h^2, b=12h-6 and c=-11, as a result, the following equations can be deduced,

(12h-6)^2 - 4*(9-3h^2)*(-11) = (144h^2 -144h +36) - (36 -12h^2)*(-11)=\\(144-(-12)*(-11)h^2 )  -144h +36 -(36*(-11)) =\\12 h^2 -144h +432 =0

therefore, dividing both sides by 12

h^2 -12 +36 =0\\(h-6)^2 =0\\h=6

8 0
2 years ago
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