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vekshin1
3 years ago
7

Which expression is equal to (f + g) (x)?

Mathematics
1 answer:
dlinn [17]3 years ago
7 0

Answer:

A

Step-by-step explanation:

(f+g)(x) means to do f(x)+g(x). We will add both functions together.

f(x)+g(x)= (x^2+3)+(x-1)=x^2+x+2

Answer a

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How do u work this out and what's the answer
I am Lyosha [343]
1) you would need to turn both fractions so they have a common denominator, that would be 2 2/4 and 3/4
2) then subtract : 2 2/4 - 3/4, and you would subtract numerators and the whole number, but you keep the denominator the same. However, you cannot subtract 2 and 3 in this case, so you need to change 2 2/4 to 1 5/4 (they are still equivalent)
3) 1 5/4 - 3/4 = 1 2/4, which simplified version is 1 1/2
Therefore, the answer is 1 and 1/2
6 0
3 years ago
Use the following trinomial to answer the questions. x^2-6x-27 When going through the process of determining the factors, the tw
Sauron [17]

Answer:

The factors are 3 and -9

Step-by-step explanation:

The given trinomial is x^2-6x-27.

We need to find two factors. We know that,

3(9) = 27

So,

x^2-6x-27 can be written as :

x^2+3x-9x-27 as two factors add up to get -6 and multiply to get -27.

Hence, the factors are 3 and -9.

5 0
2 years ago
Can you guys help ?
kobusy [5.1K]
The answer is .52
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3 0
2 years ago
lim x rightarrow 0 1 - cos ( x2 ) / 1 - cosx The limit has to be evaluated without using l'Hospital'sRule.
zaharov [31]

Answer with Step-by-step explanation:

Given

f(x)=\frac{1-cos(2x)}{1-cos(x)}\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-(cos^2{x}-sin^2{x})}{1-cos(x)})\\\\(\because cos(2x)=cos^2x-sin^2x)\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-cos^2x}{1-cos(x)}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}(\frac{(1-cosx)(1+cosx)}{1-cosx}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}((1+cosx)+\frac{sin^2x}{1-cosx})\\\\\therefore \lim_{x \rightarrow 0}f(x)=1

6 0
3 years ago
Write the equation of the quadratic function whose graph passes through <img src="https://tex.z-dn.net/?f=%28-3%2C2%29" id="TexF
blagie [28]

Answer:

f(x)=x^2+3x+2

Step-by-step explanation:

We want to write the equation of a quadratic whose graph passes through (-3, 2), (-1, 0), and (1, 6).

Remember that the standard quadratic function is given by:

f(x)=ax^2+bx+c

Since it passes through the point (-3, 2). This means that when x=-3, f(x)=f(-3)=2. Hence:

f(-3)=2=a(-3)^2+b(-3)+c

Simplify:

2=9a-3b+c

Perform the same computations for the coordinates (-1, 0) and (1, 6). Therefore:

0=a(-1)^2+b(-1)+c \\ \\0=a-b+c

And for (1, 6):

6=a(1)^2+b(1)+c\\\\ 6=a+b+c

So, we have a triple system of equations:

\left\{        \begin{array}{ll}            2=9a-3b+c &\\           0=a-b+c \\6=a+b+c        \end{array}    \right.

We can solve this using elimination.

Notice that the b term in Equation 2 and 3 are opposites. Hence, let's add them together. This yields:

(0+6)=(a+a)+(-b+b)+(c+c)

Compute:

6=2a+2c

Let's divide both sides by 2:

3=a+c

Now, let's eliminate b again but we will use Equation 1 and 2.

Notice that if we multiply Equation 2 by -3, then the b terms will be opposites. So:

-3(0)=-3(a-b+c)

Multiply:

0=-3a+3b-3c

Add this to Equation 1:

(0+2)=(9a-3a)+(-3b+3b)+(c-3c)

Compute:

2=6a-2c

Again, we can divide both sides by 2:

1=3a-c

So, we know have two equations with only two variables:

3=a+c\text{ and } 1=3a-c

We can solve for a using elimination since the c term are opposites of each other. Add the two equations together:

(3+1)=(a+3a)+(c-c)

Compute:

4=4a

Solve for a:

a=1

So, the value of a is 1.

Using either of the two equations, we can now find c. Let's use the first one. Hence:

3=a+c

Substitute 1 for a and solve for c:

\begin{aligned} c+(1)&=3 \\c&=2 \end{aligned}

So, the value of c is 2.

Finally, using any of the three original equations, solve for b:

We can use Equation 3. Hence:

6=a+b+c

Substitute in known values and solve for b:

6=(1)+b+(2)\\\\6=3+b\\\\b=3

Therefore, a=1, b=3, and c=2.

Hence, our quadratic function is:

f(x)=x^2+3x+2

5 0
3 years ago
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